#set document(title: "6.3 Graphing Parabolas", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 6.3#h(0.6em)Graphing Parabolas === Introduction The graph of a quadratic function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$f ( x ) = a x^(2) + b x + c$] is called a #strong[parabola]. Some parabolas are shown below. #figure(figph[two parabolas with labeled features], alt: "two parabolas with labeled features", caption: none) All these parabolas share certain features. - The graph has either a highest point (if the parabola opens downward, as in figure (a) or a lowest point (if the parabola opens upward, as in figure (b). This high or low point is called the #strong[vertex] of the graph. - The parabola is symmetric about a vertical line, called the #strong[axis of symmetry], that runs through the vertex. - The #math.equation(block: false, alt: "y")[$y$]-intercept is the point where the parabola intersects the #math.equation(block: false, alt: "y")[$y$]-axis. The graph of a quadratic function always has exactly one #math.equation(block: false, alt: "y")[$y$]-#strong[intercept]. - However, the graph may cross the #math.equation(block: false, alt: "x")[$x$]-axis at one point, at two points, or not at all. Points where the parabola intersects the #math.equation(block: false, alt: "x")[$x$]-axis are called the #math.equation(block: false, alt: "x")[$x$]-#strong[intercepts]. If there are two #math.equation(block: false, alt: "x")[$x$]-intercepts, they are equidistant from the axis of symmetry. - The values of the constants #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] determine the location and orientation of the parabola. We will begin by considering each of these constants separately. Which point on a parabola always lies on the axis of symmetry? \_\_\_\_\_ #solutionbox[ The vertex ] Which point on a parabola always lies on the axis of symmetry? + The #math.equation(block: false, alt: "x")[$x$]-intercept + The #math.equation(block: false, alt: "y")[$y$]-intercept + The vertex + The origin === The Graph of #math.equation(block: false, alt: "y equals a x squared")[$y = a x^(2)$] In Modeling with Functions, we saw that the graph of #math.equation(block: false, alt: "y equals a f open parenthesis x close parenthesis")[$y = a f ( x )$] is a transformation of the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$]. The scale factor, #math.equation(block: false, alt: "a")[$a$], stretches or compresses the graph vertically, and if #math.equation(block: false, alt: "a")[$a$] is negative, the graph is reflected about the #math.equation(block: false, alt: "x")[$x$]-axis. #examplebox("Example 1")[][ Sketch a graph of each quadratic function by hand. + #math.equation(block: false, alt: "y equals 2 x squared")[$y = 2 x^(2)$] + #math.equation(block: false, alt: "y equals minus the fraction 1 over 2 x squared")[$y = − display(frac(1, 2)) x^(2)$] #solutionbox[ Both functions are of the form #math.equation(block: false, alt: "y equals a x squared")[$y = a x^(2)$]. The graph of #math.equation(block: false, alt: "y equals 2 x squared")[$y = 2 x^(2)$] opens upward because #math.equation(block: false, alt: "a equals 2 greater than 0")[$a = 2 > 0$], and the graph of #math.equation(block: false, alt: "y equals minus the fraction 1 over 2 x squared")[$y = − display(frac(1, 2)) x^(2)$] opens downward because #math.equation(block: false, alt: "a equals minus the fraction 1 over 2 less than 0")[$a = − display(frac(1, 2)) < 0$]. To make a reasonable sketch by hand, it is enough to plot a few #emph[guidepoints]; the points with #math.equation(block: false, alt: "x")[$x$]-coordinates #math.equation(block: false, alt: "1")[$1$] and #math.equation(block: false, alt: "minus 1")[$− 1$] are easy to compute. #figure(figph[two vertically scaled parabolas], alt: "two vertically scaled parabolas", caption: none) #figure(table( columns: 3, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y equals 2 x squared")[$y = 2 x^(2)$]], [#math.equation(block: false, alt: "y equals minus the fraction 1 over 2 x squared")[$y = − frac(1, 2) x^(2)$]]), [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "minus the fraction 1 over 2")[$− display(frac(1, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "minus the fraction 1 over 2")[$− display(frac(1, 2))$]], )) We sketch parabolas through each set of guidepoints, as shown at left. ] ] What does the value of #math.equation(block: false, alt: "a")[$a$] tell us about the graph of #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c$] ? \_\_\_\_\_ #solutionbox[ The width of the parabola ] What does the value of #math.equation(block: false, alt: "a")[$a$] tell us about the graph of #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c$] ? + The #math.equation(block: false, alt: "y")[$y$]-intercept + The number of #math.equation(block: false, alt: "x")[$x$]-intercepts + The #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex + The width of the parabola #figure(figph[basic parabola and four vertically scaled parabolas], alt: "basic parabola and four vertically scaled parabolas", caption: none) Match each parabola in the figure above with its equation. The basic parabola is shown in black. + #math.equation(block: false, alt: "y equals minus the fraction 3 over 4 x squared")[$y = − display(frac(3, 4)) x^(2)$]\_\_\_\_\_ + #math.equation(block: false, alt: "y equals the fraction 1 over 4 x squared")[$y = display(frac(1, 4)) x^(2)$]\_\_\_\_\_ + #math.equation(block: false, alt: "y equals the fraction 5 over 2 x squared")[$y = display(frac(5, 2)) x^(2)$]\_\_\_\_\_ + #math.equation(block: false, alt: "y equals minus the fraction 5 over 4 x squared")[$y = − display(frac(5, 4)) x^(2)$]\_\_\_\_\_ #solutionbox[ + III + II + I + IV ] #figure(figph[basic parabola and four vertically scaled parabolas], alt: "basic parabola and four vertically scaled parabolas", caption: none) Match each parabola in the figure above with its equation. The basic parabola is shown in black. + #math.equation(block: false, alt: "y equals minus the fraction 3 over 4 x squared")[$y = − display(frac(3, 4)) x^(2)$] + #math.equation(block: false, alt: "y equals the fraction 1 over 4 x squared")[$y = display(frac(1, 4)) x^(2)$] + #math.equation(block: false, alt: "y equals the fraction 5 over 2 x squared")[$y = display(frac(5, 2)) x^(2)$] + #math.equation(block: false, alt: "y equals minus the fraction 5 over 4 x squared")[$y = − display(frac(5, 4)) x^(2)$] #solutionbox[ + III + II + I + IV ] === The Graph of #math.equation(block: false, alt: "y equals x squared plus c")[$y = x^(2) + c$] Next, we consider the effect of the constant term, #math.equation(block: false, alt: "c")[$c$], on the graph. Adding a constant #math.equation(block: false, alt: "c")[$c$] to the formula for #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] causes a vertical translation of the graph. #examplebox("Example 2")[][ Sketch graphs for the following quadratic functions. + #math.equation(block: false, alt: "y equals x squared minus 2")[$y = x^(2) − 2$] + #math.equation(block: false, alt: "y equals minus x squared plus 4")[$y = − x^(2) + 4$] #solutionbox[ + The graph of #math.equation(block: false, alt: "y equals x squared minus 2")[$y = x^(2) − 2$] is shifted downward by two units, compared to the basic parabola. The vertex is the point #math.equation(block: false, alt: "open parenthesis 0 , minus 2 close parenthesis")[$( 0 , − 2 )$] and the #math.equation(block: false, alt: "x")[$x$]-intercepts are the solutions of the equation #math.equation(block: true, alt: "0 equals x squared minus 2")[$0 = x^(2) − 2$] or #math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$] and #math.equation(block: false, alt: "minus the square root of 2")[$− sqrt(2)$]. The graph is shown below.#figure(figph[two vertically shifted parabolas], alt: "two vertically shifted parabolas", caption: none) + The graph of #math.equation(block: false, alt: "y equals minus x squared plus 4")[$y = − x^(2) + 4$] opens downward and is shifted #math.equation(block: false, alt: "4")[$4$] units up, compared to the basic parabola. Its vertex is the point #math.equation(block: false, alt: "open parenthesis 0 , 4 close parenthesis")[$( 0 , 4 )$]. Its #math.equation(block: false, alt: "x")[$x$]-intercepts are the solutions of the equation #math.equation(block: true, alt: "0 equals minus x squared plus 4")[$0 = − x^(2) + 4$] or #math.equation(block: false, alt: "2")[$2$] and #math.equation(block: false, alt: "minus 2")[$− 2$]. You can verify both graphs with your graphing calculator. ] ] Which of these is the equation of a parabola that has no #math.equation(block: false, alt: "x")[$x$]-intercepts? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "y equals x squared plus 4")[$y = x^(2) + 4$] ] Which of these is the equation of a parabola that has no #math.equation(block: false, alt: "x")[$x$]-intercepts? + #math.equation(block: false, alt: "y equals x squared")[$y = x^(2)$] + #math.equation(block: false, alt: "y equals x squared plus 4")[$y = x^(2) + 4$] + #math.equation(block: false, alt: "y equals x squared minus 4")[$y = x^(2) − 4$] + #math.equation(block: false, alt: "y equals x squared plus x")[$y = x^(2) + x$] #figure(figph[vertically shifted parabola], alt: "vertically shifted parabola", caption: none) + Find an equation for the parabola shown above. #linebreak() #math.equation(block: false, alt: "y equals")[$y =$]\_\_\_\_\_ + Give the #math.equation(block: false, alt: "x")[$x$]- and #math.equation(block: false, alt: "y")[$y$]-intercepts of the graph. #linebreak() #math.equation(block: false, alt: "x")[$x$]-intercepts: \_\_\_\_\_ Note: Use a comma to separate different points. #linebreak() #math.equation(block: false, alt: "y")[$y$]-intercept: \_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "y equals x squared minus 5")[$y = x^(2) − 5$] + #math.equation(block: false, alt: "x")[$x$]-intercepts: #math.equation(block: false, alt: "open parenthesis minus the square root of 5 , 0 close parenthesis , open parenthesis the square root of 5 , 0 close parenthesis")[$( − sqrt(5) , 0 ) , ( sqrt(5) , 0 )$]; #math.equation(block: false, alt: "y")[$y$]-intercept: #math.equation(block: false, alt: "open parenthesis 0 , minus 5 close parenthesis")[$( 0 , − 5 )$] ] + Find an equation for the parabola shown at right. + Give the #math.equation(block: false, alt: "x")[$x$]- and #math.equation(block: false, alt: "y")[$y$]-intercepts of the graph. #figure(figph[vertically shifted parabola], alt: "vertically shifted parabola", caption: none) #solutionbox[ + #math.equation(block: false, alt: "y equals x squared minus 5")[$y = x^(2) − 5$] + #math.equation(block: false, alt: "x")[$x$]-intercepts: #math.equation(block: false, alt: "open parenthesis minus the square root of 5 , 0 close parenthesis , open parenthesis the square root of 5 , 0 close parenthesis")[$( − sqrt(5) , 0 ) , ( sqrt(5) , 0 )$]; #math.equation(block: false, alt: "y")[$y$]-intercept: #math.equation(block: false, alt: "open parenthesis 0 , minus 5 close parenthesis")[$( 0 , − 5 )$] ] Describe what the parameters #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "c")[$c$] tell you about the graph of #math.equation(block: false, alt: "y equals a x squared plus c")[$y = a x^(2) + c$]. \_\_\_\_\_ Describe what the parameters #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "c")[$c$] tell you about the graph of #math.equation(block: false, alt: "y equals a x squared plus c")[$y = a x^(2) + c$]. === The Graph of #math.equation(block: false, alt: "y equals a x squared plus b x")[$y = a x^(2) + b x$] How does the linear term, #math.equation(block: false, alt: "b x")[$b x$], affect the graph? Let us begin by considering an example. Graph the function #math.equation(block: true, alt: "y equals 2 x squared plus 8 x")[$y = 2 x^(2) + 8 x$] on your calculator. The graph is shown at right. #figure(figph[shifted parabola], alt: "shifted parabola", caption: none) Note that #math.equation(block: false, alt: "a equals 2")[$a = 2$] and that #math.equation(block: false, alt: "2 greater than 0")[$2 > 0$], so the parabola opens upward. We can find the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph by setting #math.equation(block: false, alt: "y")[$y$] equal to zero: #math.equation(block: true, alt: "0, equals 2 x squared plus 8 x; equals 2 x open parenthesis x plus 4 close parenthesis")[$0 & = 2 x^(2) + 8 x \ & = 2 x ( x + 4 )$] The solutions of this equation are #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "minus 4")[$− 4$], so the #math.equation(block: false, alt: "x")[$x$]-intercepts are the points #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] and #math.equation(block: false, alt: "open parenthesis minus 4 , 0 close parenthesis")[$( − 4 , 0 )$]. Recall that the parabola is symmetric about a vertical line through its vertex. (We will prove that this is true in the Homework problems.) The two #math.equation(block: false, alt: "x")[$x$]-intercepts are equidistant from this line of symmetry, so the #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex lies exactly halfway between the #math.equation(block: false, alt: "x")[$x$]-intercepts. We can average their values to find #math.equation(block: true, alt: "x equals the fraction 1 over 2 open bracket 0 plus open parenthesis minus 4 close parenthesis close bracket equals minus 2")[$x = frac(1, 2) [ 0 + ( − 4 ) ] = − 2$] To find the #math.equation(block: false, alt: "y")[$y$]-coordinate of the vertex, substitute #math.equation(block: false, alt: "x equals minus 2")[$x = − 2$] into the equation for the parabola: #math.equation(block: true, alt: "y, equals 2 open parenthesis minus 2 close parenthesis squared plus 8 open parenthesis minus 2 close parenthesis; equals 8 minus 16 equals minus 8")[$y & = 2 ( − 2 )^(2) + 8 ( − 2 ) \ & = 8 − 16 = − 8$] Thus, the vertex is the point #math.equation(block: false, alt: "open parenthesis minus 2 , minus 8 close parenthesis")[$( − 2 , − 8 )$]. + Find the #math.equation(block: false, alt: "x")[$x$]-intercepts and the vertex of the parabola #math.equation(block: false, alt: "y equals 6 x minus x squared")[$y = 6 x − x^(2)$]. #linebreak() #math.equation(block: false, alt: "x")[$x$]-intercepts: \_\_\_\_\_ Note: Use a comma to separate different points. #linebreak() Vertex: \_\_\_\_\_ + Verify your answers by graphing the function in the window #math.equation(block: true, alt: "Xmin, equals minus 9.4, Xmax equals 9.4; Ymin, equals minus 10, Ymax equals 10")[$"Xmin" & = − 9.4 & & "Xmax" = 9.4 \ "Ymin" & = − 10 & & "Ymax" = 10$] #solutionbox[ #math.equation(block: false, alt: "x")[$x$]-intercepts: #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 6 , 0 close parenthesis")[$( 6 , 0 )$]; vertex: #math.equation(block: false, alt: "open parenthesis 3 , 9 close parenthesis")[$( 3 , 9 )$] ] + Find the #math.equation(block: false, alt: "x")[$x$]-intercepts and the vertex of the parabola #math.equation(block: false, alt: "y equals 6 x minus x squared")[$y = 6 x − x^(2)$]. + Verify your answers by graphing the function in the window #math.equation(block: true, alt: "Xmin, equals minus 9.4, Xmax equals 9.4; Ymin, equals minus 10, Ymax equals 10")[$"Xmin" & = − 9.4 & & "Xmax" = 9.4 \ "Ymin" & = − 10 & & "Ymax" = 10$] #solutionbox[ #math.equation(block: false, alt: "x")[$x$]-intercepts: #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 6 , 0 close parenthesis")[$( 6 , 0 )$]; vertex: #math.equation(block: false, alt: "open parenthesis 3 , 9 close parenthesis")[$( 3 , 9 )$] ] === Finding the Vertex We can use the same method to find a formula for the vertex of any parabola of the form #math.equation(block: true, alt: "y equals a x squared plus b x")[$y = a x^(2) + b x$] We proceed as we did in the previous example. First, find the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph by setting #math.equation(block: false, alt: "y")[$y$] equal to zero and solving for #math.equation(block: false, alt: "x")[$x$]. #math.equation(block: true, alt: "0, equals a x squared plus b x, Factor.; equals x open parenthesis a x plus b close parenthesis")[$0 & = a x^(2) + b x & & "Factor." \ & = x ( a x + b )$] Thus, #math.equation(block: true, alt: "x, equals 0, or, a x plus b, equals 0; x, equals 0, or, x, equals the fraction minus b over a")[$x & = 0 & "or" & & a x + b & = 0 \ x & = 0 & "or" & & x & = frac(− b, a)$] #figure(figph[parabola with vertex and line of symmetry], alt: "parabola with vertex and line of symmetry", caption: none) The #math.equation(block: false, alt: "x")[$x$]-intercepts are the points #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] and #math.equation(block: false, alt: "open parenthesis the fraction minus b over a , 0 close parenthesis")[$( display(frac(− b, a)) , 0 )$]. Next, we find the #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex by taking the average of the two #math.equation(block: false, alt: "x")[$x$]-intercepts found above: #math.equation(block: true, alt: "x equals the fraction 1 over 2 open bracket 0 plus open parenthesis the fraction minus b over a close parenthesis close bracket equals the fraction minus b over 2 a")[$x = frac(1, 2) [ 0 + ( frac(− b, a) ) ] = frac(− b, 2 a)$] This gives us a formula for the #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Vertex of a Parabola] For the graph of #math.equation(block: false, alt: "y equals a x squared plus b x")[$y = a x^(2) + b x$], the #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex is #math.equation(block: true, alt: "x sub v equals the fraction minus b over 2 a")[$x_(v) = frac(− b, 2 a)$] ] Also, the axis of symmetry is the vertical line #math.equation(block: false, alt: "x equals the fraction minus b over 2 a")[$x = display(frac(− b, 2 a))$]as shown in the figure above. Finally, we find the #math.equation(block: false, alt: "y")[$y$]-coordinate of the vertex by substituting its #math.equation(block: false, alt: "x")[$x$]-coordinate into the equation for the parabola. #examplebox("Example 3")[][ + Find the vertex of the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals minus 1.8 x squared minus 16.2 x")[$f ( x ) = − 1.8 x^(2) − 16.2 x$]. + Find the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph. #solutionbox[ + The #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex is #math.equation(block: true, alt: "x sub v equals the fraction minus b over 2 a equals the fraction minus open parenthesis minus 16.2 close parenthesis over 2 open parenthesis minus 1.8 close parenthesis equals minus 4.5")[$x_(v) = frac(− b, 2 a) = frac(− ( − 16.2 ), 2 ( − 1.8 )) = − 4.5$] To find the #math.equation(block: false, alt: "y")[$y$]-coordinate of the vertex, evaluate #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] at #math.equation(block: false, alt: "x equals minus 4.5")[$x = − 4.5$]. #math.equation(block: true, alt: "y sub v equals minus 1.8 open parenthesis minus 4.5 close parenthesis squared minus 16.2 open parenthesis minus 4.5 close parenthesis equals 36.45")[$y_(v) = − 1.8 ( − 4.5 )^(2) − 16.2 ( − 4.5 ) = 36.45$] The vertex is #math.equation(block: false, alt: "open parenthesis minus 4.5 , 36.45 close parenthesis")[$( − 4.5 , 36.45 )$]. + To find the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph, set #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0")[$f ( x ) = 0$] and solve. #math.equation(block: true, alt: "minus 1.8 x squared minus 16.2 x, equals 0, Factor.; minus x open parenthesis 1.8 x plus 16.2 close parenthesis, equals 0, Set each factor equal to zero.; minus x equals 01.8 x plus 16.2, equals 0, Solve each equation.; x equals 0 x, equals minus 9")[$− 1.8 x^(2) − 16.2 x & = 0 & & "Factor." \ − x ( 1.8 x + 16.2 ) & = 0 & & "Set each factor equal to zero." \ − x = 0 #hide($b l a n k$) 1.8 x + 16.2 & = 0 & & "Solve each equation." \ x = 0 #hide($b l a n k b l a n k b l a n$) x & = − 9$] The #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph are #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] and #math.equation(block: false, alt: "open parenthesis minus 9 , 0 close parenthesis")[$( − 9 , 0 )$]. The graph is shown below. #figure(figph[downward opening parabola], alt: "downward opening parabola", caption: none) ] ] Explain why the #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex is the average of the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph. \_\_\_\_\_ Explain why the #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex is the average of the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph. === The Graph of #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c$] Now we will see that the vertex formula holds for any parabola. Consider the function #math.equation(block: true, alt: "y equals 2 x squared plus 8 x plus 6")[$y = 2 x^(2) + 8 x + 6$] Adding #math.equation(block: false, alt: "6")[$6$] to #math.equation(block: false, alt: "2 x squared plus 8 x")[$2 x^(2) + 8 x$] shifts each point on the graph #math.equation(block: false, alt: "6")[$6$] units upward, as shown at right. The #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex will not be affected by an upward shift. Thus, the formula #math.equation(block: true, alt: "x sub v equals the fraction minus b over 2 a")[$x_(v) = frac(− b, 2 a)$] for the #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex still holds. We have #math.equation(block: true, alt: "x sub v equals the fraction minus 8 over 2 open parenthesis 2 close parenthesis equals minus 2")[$x_(v) = frac(− 8, 2 ( 2 )) = − 2$] We find the #math.equation(block: false, alt: "y")[$y$]-coordinate of the vertex by substituting #math.equation(block: false, alt: "x sub v equals minus 2")[$x_(v) = − 2$] into the equation for the parabola. #figure(figph[Two upward-opening parabolas on a coordinate grid. The dashed blue graph of y = 2x squared + 8x has its vertex below the x-axis; the solid red graph of y = 2x squared + 8x + 6 is that parabola shifted up 6 units, passing through the points (negative 4, 6) and (0, 6).], alt: "Two upward-opening parabolas on a coordinate grid. The dashed blue graph of y = 2x squared + 8x has its vertex below the x-axis; the solid red graph of y = 2x squared + 8x + 6 is that parabola shifted up 6 units, passing through the points (negative 4, 6) and (0, 6).", caption: none) #math.equation(block: true, alt: "y sub v, equals 2 open parenthesis minus 2 close parenthesis squared plus 8 open parenthesis minus 2 close parenthesis plus 6; equals 8 minus 16 plus 6 equals minus 2")[$y_(v) & = 2 ( − 2 )^(2) + 8 ( − 2 ) + 6 \ & = 8 − 16 + 6 = − 2$] So the vertex is the point #math.equation(block: false, alt: "open parenthesis minus 2 , minus 2 close parenthesis")[$( − 2 , − 2 )$]. (Notice that this point is shifted #math.equation(block: false, alt: "6")[$6$] units upward from the vertex of #math.equation(block: false, alt: "y equals 2 x squared plus 8 x")[$y = 2 x^(2) + 8 x$].) We find the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph by setting #math.equation(block: false, alt: "y")[$y$] equal to zero. #math.equation(block: true, alt: "0, equals 2 x squared plus 8 x plus 6, Factor the right side.; equals 2 open parenthesis x plus 1 close parenthesis open parenthesis x plus 3 close parenthesis, Set each factor equal to zero.; x plus 1, equals 0 or x plus 3 equals 0; x, equals minus 1 x equals minus 3")[$0 & = 2 x^(2) + 8 x + 6 & & "Factor the right side." \ & = 2 ( x + 1 ) ( x + 3 ) & & "Set each factor equal to zero." \ x + 1 & = 0 " " " " " " " or " " " " " " " x + 3 = 0 \ x & = − 1 #hide($b l a n k b l$) x = − 3$] The #math.equation(block: false, alt: "x")[$x$]-intercepts are the points #math.equation(block: false, alt: "open parenthesis minus 1 , 0 close parenthesis")[$( − 1 , 0 )$] and #math.equation(block: false, alt: "open parenthesis minus 3 , 0 close parenthesis")[$( − 3 , 0 )$]. The #math.equation(block: false, alt: "y")[$y$]-intercept of the graph is found by setting #math.equation(block: false, alt: "x")[$x$] equal to zero: #math.equation(block: true, alt: "y equals 2 open parenthesis 0 close parenthesis 2 plus 8 open parenthesis 0 close parenthesis plus 6 equals 6")[$y = 2 ( 0 ) 2 + 8 ( 0 ) + 6 = 6$] You can see that the #math.equation(block: false, alt: "y")[$y$]-intercept, #math.equation(block: false, alt: "6")[$6$], is just the constant term of the quadratic equation. The completed graph is shown above. The #math.equation(block: false, alt: "y")[$y$]-intercept of #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c$] is the same as the value of which parameter? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "c")[$c$] ] The #math.equation(block: false, alt: "y")[$y$]-intercept of #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c$] is the same as the value of which parameter? + #math.equation(block: false, alt: "a")[$a$] + #math.equation(block: false, alt: "b")[$b$] + #math.equation(block: false, alt: "c")[$c$] + None of these #examplebox("Example 4")[][ Find the vertex of the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals minus 2 x squared plus x plus 1")[$f ( x ) = − 2 x^(2) + x + 1$]. #solutionbox[ For this function, #math.equation(block: false, alt: "a equals minus 2")[$a = − 2$], #math.equation(block: false, alt: "b equals 1")[$b = 1$], and #math.equation(block: false, alt: "c equals 1")[$c = 1$]. The #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex is given by #math.equation(block: true, alt: "x sub v equals the fraction minus b over 2 a equals the fraction minus 1 over 2 open parenthesis minus 2 close parenthesis equals the fraction 1 over 4")[$x_(v) = frac(− b, 2 a) = frac(− 1, 2 ( − 2 )) = frac(1, 4)$] To find the #math.equation(block: false, alt: "y")[$y$]-coordinate of the vertex, we substitute #math.equation(block: false, alt: "x equals the fraction 1 over 4")[$x = display(frac(1, 4))$] into the equation. We can do this by hand to find #math.equation(block: true, alt: "y sub v, equals minus 2 open parenthesis the fraction 1 over 4 close parenthesis squared plus the fraction 1 over 4 plus 1; equals minus 2 open parenthesis the fraction 1 over 16 close parenthesis plus the fraction 4 over 16 plus the fraction 16 over 16 equals the fraction 18 over 16 equals the fraction 9 over 8")[$y_(v) & = − 2 attach(( frac(1, 4) ), t: 2) + frac(1, 4) + 1 \ & = − 2 ( frac(1, 16) ) + frac(4, 16) + frac(16, 16) = frac(18, 16) = frac(9, 8)$] So the coordinates of the vertex are #math.equation(block: false, alt: "open parenthesis the fraction 1 over 4 , the fraction 9 over 8 close parenthesis")[$( display(frac(1, 4)) , display(frac(9, 8)) )$]. Alternatively, we can use the calculator to evaluate #math.equation(block: false, alt: "minus 2 x squared plus x plus 1")[$− 2 x^(2) + x + 1$] for #math.equation(block: false, alt: "x equals 0.25")[$x = 0.25$]. The calculator returns the #math.equation(block: false, alt: "y")[$y$]-value #math.equation(block: false, alt: "1.125")[$1.125$]. Thus, the vertex is the point #math.equation(block: false, alt: "open parenthesis 0.25 , 1.125 close parenthesis")[$( 0.25 , 1.125 )$], which is the decimal equivalent of #math.equation(block: false, alt: "open parenthesis the fraction 1 over 4 , the fraction 9 over 8 close parenthesis")[$( display(frac(1, 4)) , display(frac(9, 8)) )$]. ] ] Find the vertex of the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x squared minus 6 x plus 4")[$f ( x ) = 3 x^(2) − 6 x + 4$]. \_\_\_\_\_ Decide whether the vertex is a maximum point or a minimum point of the graph.\_\_\_\_\_ #solutionbox[ #math.equation(block: false, alt: "open parenthesis 1 , 1 close parenthesis")[$( 1 , 1 )$], minimum ] Find the vertex of the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x squared minus 6 x plus 4")[$f ( x ) = 3 x^(2) − 6 x + 4$]. Decide whether the vertex is a maximum point or a minimum point of the graph. #solutionbox[ #math.equation(block: false, alt: "open parenthesis 1 , 1 close parenthesis")[$( 1 , 1 )$], minimum ] === Number of #math.equation(block: false, alt: "x")[$x$]-Intercepts The graph of the quadratic function #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$f ( x ) = a x^(2) + b x + c$] may have two, one, or no #math.equation(block: false, alt: "x")[$x$]-intercepts, according to the number of distinct real-valued solutions of the equation #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$]. Consider the three functions graphed below. #figure(figph[parabolas with zero, one, or two x-intercepts], alt: "parabolas with zero, one, or two x-intercepts", caption: none) - The graph of #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x squared minus 4 x plus 3")[$f ( x ) = x^(2) − 4 x + 3$] has two #math.equation(block: false, alt: "x")[$x$]-intercepts, because the equation #math.equation(block: true, alt: "x squared minus 4 x plus 3 equals 0")[$x^(2) − 4 x + 3 = 0$] has two real-valued solutions, #math.equation(block: false, alt: "x equals 1")[$x = 1$] and #math.equation(block: false, alt: "x equals 3")[$x = 3$]. - The graph of #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals x squared minus 4 x plus 4")[$g ( x ) = x^(2) − 4 x + 4$] has only one #math.equation(block: false, alt: "x")[$x$]-intercept, because the equation #math.equation(block: true, alt: "x squared minus 4 x plus 4 equals 0")[$x^(2) − 4 x + 4 = 0$] has only one (repeated) real-valued solution, #math.equation(block: false, alt: "x equals 2")[$x = 2$]. - The graph of #math.equation(block: true, alt: "h open parenthesis x close parenthesis equals x squared minus 4 x plus 6")[$h ( x ) = x^(2) − 4 x + 6$] has no #math.equation(block: false, alt: "x")[$x$]-intercepts, because the equation #math.equation(block: true, alt: "x squared minus 4 x plus 6 equals 0")[$x^(2) − 4 x + 6 = 0$] has no real-valued solutions. A closer look at the quadratic formula reveals useful information about the solutions of quadratic equations. For the three functions above, we have the following: #figure(figph[discriminant for the three cases], alt: "discriminant for the three cases", caption: none) The expression #math.equation(block: false, alt: "b squared minus 4 a c")[$b^(2) − 4 a c$], which appears under the radical in the quadratic formula, is called the #strong[discriminant], #math.equation(block: false, alt: "D")[$D$], of the equation. The value of the discriminant determines the nature of the solutions of the equation. In particular, if the discriminant is negative, the solutions of the quadratic equation are #strong[complex numbers]. (We will study complex numbers in Complex Numbers.) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The Discriminant] The #strong[discriminant] of a quadratic equation is #math.equation(block: false, alt: "D equals b squared minus 4 a c")[$D = b^(2) − 4 a c$]. + If #math.equation(block: false, alt: "D greater than 0")[$D > 0$], there are two unequal real solutions. + If #math.equation(block: false, alt: "D equals 0")[$D = 0$], there is one real solution of multiplicity two. + If #math.equation(block: false, alt: "D less than 0")[$D < 0$], there are two complex solutions. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ We can also use the discriminant to decide whether a quadratic equation can be solved by factoring. First, clear the equation of fractions. If the discriminant is a perfect square, that is, the square of an integer, the solutions are rational numbers. This in turn means that the equation can be solved by factoring. If the discriminant is not a perfect square, the solutions will be irrational. Irrational solutions always occur in conjugate pairs, #math.equation(block: true, alt: "the fraction minus b plus the square root of b squared minus 4 a c over 2 a and the fraction minus b minus the square root of b squared minus 4 a c over 2 a")[$frac(− b + sqrt(b^(2) − 4 a c), 2 a) " " " " " " "and" " " " " " " frac(− b − sqrt(b^(2) − 4 a c), 2 a)$] The only difference between the two solutions is the sign between the terms. For example, if we know that one solution of a particular quadratic equation is #math.equation(block: false, alt: "3 plus the square root of 2")[$3 + sqrt(2)$], then the other solution must be #math.equation(block: false, alt: "3 minus the square root of 2")[$3 − sqrt(2)$]. ] #examplebox("Example 5")[][ Use the discriminant to determine the nature of the solutions of each equation. Can the equation be solved by factoring? + #math.equation(block: false, alt: "x squared minus x minus 3 equals 0")[$x^(2) − x − 3 = 0$] + #math.equation(block: false, alt: "2 x squared plus x plus 1 equals 0")[$2 x^(2) + x + 1 = 0$] + #math.equation(block: false, alt: "x squared minus 2 x minus 3 equals 0")[$x^(2) − 2 x − 3 = 0$] #solutionbox[ + #math.equation(block: false, alt: "D equals b squared minus 4 a c equals open parenthesis minus 1 close parenthesis squared minus 4 open parenthesis 1 close parenthesis open parenthesis minus 3 close parenthesis equals 13 greater than 0")[$D = b^(2) − 4 a c = ( − 1 )^(2) − 4 ( 1 ) ( − 3 ) = 13 > 0$]. #linebreak() The equation has two real, unequal solutions. Because #math.equation(block: false, alt: "13")[$13$] is not a perfect square, the solutions will be irrational numbers, so the equation cannot be solved by factoring. + #math.equation(block: false, alt: "D equals b squared minus 4 a c equals 1 squared minus 4 open parenthesis 2 close parenthesis open parenthesis 1 close parenthesis equals minus 7 less than 0")[$D = b^(2) − 4 a c = 1^(2) − 4 ( 2 ) ( 1 ) = − 7 < 0$]. #linebreak() The equation has two complex solutions, which cannot be found by factoring. + #math.equation(block: false, alt: "D equals b squared minus 4 a c equals open parenthesis minus 2 close parenthesis squared minus 4 open parenthesis 1 close parenthesis open parenthesis minus 3 close parenthesis equals 16 greater than 0")[$D = b^(2) − 4 a c = ( − 2 )^(2) − 4 ( 1 ) ( − 3 ) = 16 > 0$]. #linebreak() The equation has two real, unequal solutions. Because #math.equation(block: false, alt: "16 equals 4 squared")[$16 = 4^(2)$], the solutions are rational numbers and can be found by factoring. (You can verify the conclusions above by solving each equation.) ] ] Use the discriminant to discover how many #math.equation(block: false, alt: "x")[$x$]-intercepts the graph of each function has. + #math.equation(block: false, alt: "y equals x squared plus 5 x plus 7")[$y = x^(2) + 5 x + 7$] \_\_\_\_\_ + #math.equation(block: false, alt: "y equals minus the fraction 1 over 2 x squared plus 4 x minus 8")[$y = − display(frac(1, 2)) x^(2) + 4 x − 8$] \_\_\_\_\_ #solutionbox[ + None: the discriminant is negative. + One: the discriminant is 0. ] Use the discriminant to discover how many #math.equation(block: false, alt: "x")[$x$]-intercepts the graph of each function has. + #math.equation(block: false, alt: "y equals x squared plus 5 x plus 7")[$y = x^(2) + 5 x + 7$] + #math.equation(block: false, alt: "y equals minus the fraction 1 over 2 x squared plus 4 x minus 8")[$y = − display(frac(1, 2)) x^(2) + 4 x − 8$] #solutionbox[ + None: the discriminant is negative. + One: the discriminant is 0. ] In Practice 5, you can check that the single #math.equation(block: false, alt: "x")[$x$]-intercept is also the vertex of the parabola. In Practice 5, you can check that the single #math.equation(block: false, alt: "x")[$x$]-intercept is also the vertex of the parabola. Explain what the discriminant tells us about a quadratic equation. \_\_\_\_\_ Explain what the discriminant tells us about a quadratic equation. === Sketching a Parabola Once we have located the vertex of the parabola, the #math.equation(block: false, alt: "x")[$x$]-intercepts, and the #math.equation(block: false, alt: "y")[$y$]-intercept, we can sketch a reasonably accurate graph. Recall that the graph should be symmetric about a vertical line through the vertex. We summarize the procedure as follows. The sketching recipe's two landmarks, live: y = a(x − h)² + k with the vertex marked in red at (h, k) and the axis of symmetry dashed at x = h. Slide h and k and the whole parabola follows its vertex; the graph is always the mirror image of itself across the dashed line, which is why finding the vertex is most of the sketch. The slider a controls the rest: its sign says whether the parabola opens upward (a \> 0, vertex a minimum) or downward (a \< 0, vertex a maximum), and its size says how narrow the bowl is. At a = 0 the parabola flattens into the horizontal line through its vertex — the boundary between the two cases.{"functions":\[{"color":"\#1f4e79","expression":"a\*(x - h)^2 + k","expression2":"","id":"parabola","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true},{"color":"\#6a6a6a","expression":"h + 0\*t","expression2":"t","id":"axis-of-symmetry","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"parametric","tMax":6,"tMin":-6,"variable":"x","visible":true},{"color":"\#b8442a","expression":"h + 0.12\*cos(t)","expression2":"k + 0.12\*sin(t)","id":"vertex-marker","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":6.2832,"tMin":0,"variable":"x","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-a","label":"Opens/width a","max":3,"min":-3,"name":"a","step":0.25,"value":1},{"id":"p-h","label":"Vertex x: h","max":3,"min":-3,"name":"h","step":0.1,"value":1},{"id":"p-k","label":"Vertex y: k","max":4,"min":-4,"name":"k","step":0.1,"value":-2}\],"title":"y = a(x \\u2212 h)\\u00b2 + k: vertex (h, k), axis x = h","version":1,"viewport":{"centerX":0,"centerY":0,"scale":44}}#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[To Graph the Quadratic Function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$f ( x ) = a x^(2) + b x + c$]:] + Determine whether the parabola opens upward (if #math.equation(block: false, alt: "a greater than 0")[$a > 0$]) or downward (if #math.equation(block: false, alt: "a less than 0")[$a < 0$]). + Locate the vertex of the parabola. + The #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex is #math.equation(block: false, alt: "x sub v equals the fraction minus b over 2 a")[$x_(v) = display(frac(− b, 2 a))$]. + Find the #math.equation(block: false, alt: "y")[$y$]-coordinate of the vertex by substituting #math.equation(block: false, alt: "x sub v")[$x_(v)$] into the equation of the parabola. + Locate the #math.equation(block: false, alt: "x")[$x$]-intercepts (if any) by setting #math.equation(block: false, alt: "y equals 0")[$y = 0$] and solving for #math.equation(block: false, alt: "x")[$x$]. + Locate the #math.equation(block: false, alt: "y")[$y$]-intercept by evaluating #math.equation(block: false, alt: "y")[$y$] for #math.equation(block: false, alt: "x equals 0")[$x = 0$]. + Locate the point symmetric to the #math.equation(block: false, alt: "y")[$y$]-intercept across the axis of symmetry. ] #examplebox("Example 6")[][ Sketch a graph of the equation #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared plus 3 x plus 1")[$f ( x ) = x^(2) + 3 x + 1$], showing the significant points. #solutionbox[ We follow the steps outlined above. + Because #math.equation(block: false, alt: "a equals 1 greater than 0")[$a = 1 > 0$], we know that the parabola opens upward. + We compute the coordinates of the vertex: #math.equation(block: true, alt: "x sub v, equals the fraction minus b over 2 a equals the fraction minus 3 over 2 open parenthesis 1 close parenthesis equals minus 1.5; y sub v, equals open parenthesis minus 1.5 close parenthesis 2 plus 3 open parenthesis minus 1.5 close parenthesis plus 1 equals minus 1.25")[$x_(v) & = frac(− b, 2 a) = frac(− 3, 2 ( 1 )) = − 1.5 \ y_(v) & = ( − 1.5 ) 2 + 3 ( − 1.5 ) + 1 = − 1.25$] The vertex is the point #math.equation(block: false, alt: "open parenthesis minus 1.5 , minus 1.25 close parenthesis")[$( − 1.5 , − 1.25 )$]. + We set #math.equation(block: false, alt: "y")[$y$] equal to zero to find the #math.equation(block: false, alt: "x")[$x$]-intercepts. #math.equation(block: true, alt: "0, equals x squared plus 3 x plus 1, Use the quadratic formula.; x, equals the fraction minus 3 plus or minus the square root of 3 squared minus 4 open parenthesis 1 close parenthesis open parenthesis 1 close parenthesis over 2 open parenthesis 1 close parenthesis; equals the fraction minus 3 plus or minus the square root of 5 over 2")[$0 & = x^(2) + 3 x + 1 & & "Use the quadratic formula." \ x & = frac(− 3 ± sqrt(3^(2) − 4 ( 1 ) ( 1 )), 2 ( 1 )) \ & = frac(− 3 ± sqrt(5), 2)$] Rounding to the nearest tenth, we find that the #math.equation(block: false, alt: "x")[$x$]-intercepts are approximately #math.equation(block: false, alt: "open parenthesis minus 2.6 , 0 close parenthesis")[$( − 2.6 , 0 )$] and #math.equation(block: false, alt: "open parenthesis minus 0.4 , 0 close parenthesis")[$( − 0.4 , 0 )$]. + We substitute #math.equation(block: false, alt: "x equals 0")[$x = 0$] to find the #math.equation(block: false, alt: "y")[$y$]-intercept, #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$]. + The axis of symmetry is the vertical line #math.equation(block: false, alt: "x equals minus 1.5")[$x = − 1.5$], so the #math.equation(block: false, alt: "y")[$y$]-intercept lies #math.equation(block: false, alt: "1.5")[$1.5$] units to the right of the axis of symmetry. #linebreak() There must be another point on the parabola with the same #math.equation(block: false, alt: "y")[$y$]-coordinate as the #math.equation(block: false, alt: "y")[$y$]-intercept but #math.equation(block: false, alt: "1.5")[$1.5$] units to the left of the axis of symmetry. The coordinates of this point are #math.equation(block: false, alt: "open parenthesis minus 3 , 1 close parenthesis")[$( − 3 , 1 )$]. #linebreak() Finally, plot the #math.equation(block: false, alt: "x")[$x$]-intercepts, the vertex, and the #math.equation(block: false, alt: "y")[$y$]-intercept and its symmetric point, and draw a parabola through them. The finished graph is shown below.#figure(figph[parabola], alt: "parabola", caption: none) ] ] + Find the intercepts and the vertex of the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 5 x plus 4")[$f ( x ) = x^(2) − 5 x + 4$]. #linebreak() Intercepts: \_\_\_\_\_ Note: Use a comma to separate different points. #linebreak() Vertex: \_\_\_\_\_ + Sketch the graph by hand. + Use your calculator to verify your graph. #solutionbox[ + #math.equation(block: false, alt: "open parenthesis 0 , 4 close parenthesis")[$( 0 , 4 )$]; #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$], #math.equation(block: false, alt: "open parenthesis 4 , 0 close parenthesis")[$( 4 , 0 )$]; vertex #math.equation(block: false, alt: "open parenthesis the fraction 5 over 2 , the fraction minus 9 over 4 close parenthesis")[$( display(frac(5, 2)) , display(frac(− 9, 4)) )$] + A graph is below. ] A graph for part (b): #figure(figph[parabola], alt: "parabola", caption: none) + Find the intercepts and the vertex of the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 5 x plus 4")[$f ( x ) = x^(2) − 5 x + 4$]. + Sketch the graph by hand. + Use your calculator to verify your graph. #solutionbox[ + #math.equation(block: false, alt: "open parenthesis 0 , 4 close parenthesis")[$( 0 , 4 )$]; #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$], #math.equation(block: false, alt: "open parenthesis 4 , 0 close parenthesis")[$( 4 , 0 )$]; vertex #math.equation(block: false, alt: "open parenthesis the fraction 5 over 2 , the fraction minus 9 over 4 close parenthesis")[$( display(frac(5, 2)) , display(frac(− 9, 4)) )$] + #figure(figph[parabola], alt: "parabola", caption: none) ] Which points on a parabola should you find to help you sketch its graph? Describe how to find each of these points. \_\_\_\_\_ Which points on a parabola should you find to help you sketch its graph? Describe how to find each of these points. === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Vertex - Conjugate pair - Axis of symmetry ==== CONCEPTS + The graph of a quadratic function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$f ( x ) = a x^(2) + b x + c$] is called a #strong[parabola]. The values of the constants #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] determine the location and orientation of the parabola. + For the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$f ( x ) = a x^(2) + b x + c$], the #math.equation(block: false, alt: "x")[$x$]-coordinate of the #strong[vertex] is #math.equation(block: false, alt: "x sub v equals the fraction minus b over 2 a")[$x_(v) = frac(− b, 2 a)$]. To find the #math.equation(block: false, alt: "y")[$y$]-coordinate of the vertex, we substitute #math.equation(block: false, alt: "x sub v")[$x_(v)$] into the formula for the parabola. + The graph of the quadratic function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$f ( x ) = a x^(2) + b x + c$] may have two, one, or no #math.equation(block: false, alt: "x")[$x$]-intercepts, according to the number of distinct real-valued solutions of the equation #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$]. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The Discriminant] The #strong[discriminant] of a quadratic equation is #math.equation(block: false, alt: "D equals b squared minus 4 a c")[$D = b^(2) − 4 a c$]. + If #math.equation(block: false, alt: "D greater than 0")[$D > 0$], there are two unequal real solutions. + If #math.equation(block: false, alt: "D equals 0")[$D = 0$], there is one real solution of multiplicity two. + If #math.equation(block: false, alt: "D less than 0")[$D < 0$], there are two complex solutions. ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[To Graph the Quadratic Function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$f ( x ) = a x^(2) + b x + c$]:] + Determine whether the parabola opens upward (if #math.equation(block: false, alt: "a greater than 0")[$a > 0$]) or downward (if #math.equation(block: false, alt: "a less than 0")[$a < 0$]). + Locate the vertex of the parabola. + The #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex is #math.equation(block: false, alt: "x sub v equals the fraction minus b over 2 a")[$x_(v) = display(frac(− b, 2 a))$]. + Find the #math.equation(block: false, alt: "y")[$y$]-coordinate of the vertex by substituting #math.equation(block: false, alt: "x sub v")[$x_(v)$] into the equation of the parabola. + Locate the #math.equation(block: false, alt: "x")[$x$]-intercepts (if any) by setting #math.equation(block: false, alt: "y equals 0")[$y = 0$] and solving for #math.equation(block: false, alt: "x")[$x$]. + Locate the #math.equation(block: false, alt: "y")[$y$]-intercept by evaluating #math.equation(block: false, alt: "y")[$y$] for #math.equation(block: false, alt: "x equals 0")[$x = 0$]. + Locate the point symmetric to the #math.equation(block: false, alt: "y")[$y$]-intercept across the axis of symmetry. ] ==== STUDY QUESTIONS + Sketch a parabola that opens downward. Show the location of the #math.equation(block: false, alt: "x")[$x$]-intercepts, the #math.equation(block: false, alt: "y")[$y$]-intercept, the vertex, and the axis of symmetry. + Describe how the value of #math.equation(block: false, alt: "a")[$a$] in #math.equation(block: false, alt: "y equals a x squared")[$y = a x^(2)$] alters the graph of the basic parabola. + Describe how the value of #math.equation(block: false, alt: "c")[$c$] in #math.equation(block: false, alt: "y equals x squared plus c")[$y = x^(2) + c$] alters the graph of the basic parabola. + Suppose you know that the #math.equation(block: false, alt: "x")[$x$]-intercepts of a parabola are #math.equation(block: false, alt: "open parenthesis minus 8 , 0 close parenthesis")[$( − 8 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$]. What is the equation of the parabola's axis of symmetry? + State a formula for the #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex of a parabola. How can you find the #math.equation(block: false, alt: "y")[$y$]-coordinate of the vertex? + Suppose that a given parabola has only one #math.equation(block: false, alt: "x")[$x$]-intercept. What can you say about the vertex of the parabola? + Explain why a quadratic equation has one (repeated) solution if its discriminant is zero, and none if the discriminant is negative. ==== SKILLS Practice each skill in the Homework problems listed. + Graph transformations of the basic parabola: \#1 and 2, 7, and 8 + Locate the #math.equation(block: false, alt: "x")[$x$]-intercepts of a parabola: \#3–6 + Locate the vertex of a parabola: \#3–6, 13, and 14 + Sketch the graph of a quadratic function: \#15–24, 41, and 42 + Use the discriminant to describe the solutions of a quadratic equation: \#25–40 === Homework 6.3 For Problems 1–2, describe what each graph will look like compared to the basic parabola. Then sketch a graph by hand and label the coordinates of three points on the graph. + #math.equation(block: false, alt: "y equals 2 x squared")[$y = 2 x^(2)$] + #math.equation(block: false, alt: "y equals 2 plus x squared")[$y = 2 + x^(2)$] + #math.equation(block: false, alt: "y equals open parenthesis x plus 2 close parenthesis squared")[$y = ( x + 2 )^(2)$] + #math.equation(block: false, alt: "y equals x squared minus 2")[$y = x^(2) − 2$] + The parabola opens up, twice as steep as the standard parabola.#figure(figph[parabola], alt: "parabola", caption: none) + The parabola is the standard parabola shifted 2 units up.#figure(figph[parabola], alt: "parabola", caption: none) + The parabola is the standard parabola shifted 2 units left.#figure(figph[parabola], alt: "parabola", caption: none) + The parabola is the standard parabola shifted 2 units down.#figure(figph[parabola], alt: "parabola", caption: none) + #math.equation(block: false, alt: "y equals minus 4 x squared")[$y = − 4 x^(2)$] + #math.equation(block: false, alt: "y equals open parenthesis x minus 4 close parenthesis squared")[$y = ( x − 4 )^(2)$] + #math.equation(block: false, alt: "y equals 4 minus x squared")[$y = 4 − x^(2)$] + #math.equation(block: false, alt: "y equals x squared minus 4")[$y = x^(2) − 4$] For problems 3–6, find the vertex and the #math.equation(block: false, alt: "x")[$x$]-intercepts (if there are any) of the graph. Then sketch the graph by hand. + #math.equation(block: false, alt: "y equals x squared minus 16")[$y = x^(2) − 16$] + #math.equation(block: false, alt: "y equals 16 minus x squared")[$y = 16 − x^(2)$] + #math.equation(block: false, alt: "y equals 16 x minus x squared")[$y = 16 x − x^(2)$] + #math.equation(block: false, alt: "y equals x squared minus 16 x")[$y = x^(2) − 16 x$] + Vertex #math.equation(block: false, alt: "open parenthesis 0 , minus 16 close parenthesis")[$( 0 , − 16 )$]; #math.equation(block: false, alt: "x")[$x$]-intercepts #math.equation(block: false, alt: "open parenthesis plus or minus 4 , 0 close parenthesis")[$( ± 4 , 0 )$]#figure(figph[parabola], alt: "parabola", caption: none) + Vertex #math.equation(block: false, alt: "open parenthesis 0 , 16 close parenthesis")[$( 0 , 16 )$]; #math.equation(block: false, alt: "x")[$x$]-intercepts #math.equation(block: false, alt: "open parenthesis plus or minus 4 , 0 close parenthesis")[$( ± 4 , 0 )$]#figure(figph[parabola], alt: "parabola", caption: none) + Vertex #math.equation(block: false, alt: "open parenthesis 8 , 64 close parenthesis")[$( 8 , 64 )$]; #math.equation(block: false, alt: "x")[$x$]-intercepts #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 16 , 0 close parenthesis")[$( 16 , 0 )$]#figure(figph[parabola], alt: "parabola", caption: none) + Vertex #math.equation(block: false, alt: "open parenthesis 8 , minus 64 close parenthesis")[$( 8 , − 64 )$]; #math.equation(block: false, alt: "x")[$x$]-intercepts #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 16 , 0 close parenthesis")[$( 16 , 0 )$]#figure(figph[parabola], alt: "parabola", caption: none) + #math.equation(block: false, alt: "y equals x squared minus 1")[$y = x^(2) − 1$] + #math.equation(block: false, alt: "y equals 1 minus x squared")[$y = 1 − x^(2)$] + #math.equation(block: false, alt: "y equals x squared minus x")[$y = x^(2) − x$] + #math.equation(block: false, alt: "y equals x minus x squared")[$y = x − x^(2)$] + #math.equation(block: false, alt: "y equals 3 x squared plus 6 x")[$y = 3 x^(2) + 6 x$] + #math.equation(block: false, alt: "y equals 3 x squared minus 6 x")[$y = 3 x^(2) − 6 x$] + #math.equation(block: false, alt: "y equals 3 x squared plus 6")[$y = 3 x^(2) + 6$] + #math.equation(block: false, alt: "y equals 3 x squared minus 6")[$y = 3 x^(2) − 6$] + Vertex #math.equation(block: false, alt: "open parenthesis 1 , minus 3 close parenthesis")[$( 1 , − 3 )$]; #math.equation(block: false, alt: "x")[$x$]-intercepts #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] and #math.equation(block: false, alt: "open parenthesis minus 2 , 0 close parenthesis")[$( − 2 , 0 )$]#figure(figph[parabola], alt: "parabola", caption: none) + Vertex #math.equation(block: false, alt: "open parenthesis 1 , minus 3 close parenthesis")[$( 1 , − 3 )$]; #math.equation(block: false, alt: "x")[$x$]-intercepts #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$]#figure(figph[parabola], alt: "parabola", caption: none) + Vertex #math.equation(block: false, alt: "open parenthesis 0 , 6 close parenthesis")[$( 0 , 6 )$]; no #math.equation(block: false, alt: "x")[$x$]-intercepts #figure(figph[parabola], alt: "parabola", caption: none) + Vertex #math.equation(block: false, alt: "open parenthesis 0 , minus 6 close parenthesis")[$( 0 , − 6 )$]; #math.equation(block: false, alt: "x")[$x$]-intercepts #math.equation(block: false, alt: "open parenthesis plus or minus the square root of 2 , 0 close parenthesis")[$( ± sqrt(2) , 0 )$]#figure(figph[parabola], alt: "parabola", caption: none) + #math.equation(block: false, alt: "y equals 12 x minus 2 x squared")[$y = 12 x − 2 x^(2)$] + #math.equation(block: false, alt: "y equals 12 minus 2 x squared")[$y = 12 − 2 x^(2)$] + #math.equation(block: false, alt: "y equals 12 plus 2 x squared")[$y = 12 + 2 x^(2)$] + #math.equation(block: false, alt: "y equals 12 x plus 2 x squared")[$y = 12 x + 2 x^(2)$] Match each function with its graph. In each equation, #math.equation(block: false, alt: "a greater than 0")[$a > 0$]. + #math.equation(block: false, alt: "y equals x squared plus a")[$y = x^(2) + a$] + #math.equation(block: false, alt: "y equals x squared plus a x")[$y = x^(2) + a x$] + #math.equation(block: false, alt: "y equals a x squared")[$y = a x^(2)$] + #math.equation(block: false, alt: "y equals a x")[$y = a x$] + #math.equation(block: false, alt: "y equals x plus a")[$y = x + a$] + #math.equation(block: false, alt: "y equals x squared minus a")[$y = x^(2) − a$] #figure(figph[six graphs], alt: "six graphs", caption: none) + II + IV + I + III + VI + V Match each function with its graph. In each equation, #math.equation(block: false, alt: "b greater than 0")[$b > 0$]. + #math.equation(block: false, alt: "y equals minus b x")[$y = − b x$] + #math.equation(block: false, alt: "y equals minus b x squared")[$y = − b x^(2)$] + #math.equation(block: false, alt: "y equals b minus x squared")[$y = b − x^(2)$] + #math.equation(block: false, alt: "y equals x minus b")[$y = x − b$] + #math.equation(block: false, alt: "y equals b minus x")[$y = b − x$] + #math.equation(block: false, alt: "y equals x squared minus b x")[$y = x^(2) − b x$] #figure(figph[six graphs], alt: "six graphs", caption: none) Commercial fishermen rely on a steady supply of fish in their area. To avoid overfishing, they adjust their harvest to the size of the population. The function #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals 0.4 x minus 0.0001 x squared")[$g ( x ) = 0.4 x − 0.0001 x^(2)$] gives the annual rate of growth, in tons per year, of a fish population of biomass #math.equation(block: false, alt: "x")[$x$] tons. + Find the vertex of the graph. What does it tell us about the fish population? + Sketch the graph for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 5000")[$0 ≤ x ≤ 5000$]. + For what values of #math.equation(block: false, alt: "x")[$x$] does the fish population decrease rather than increase? Suggest a reason why the population might decrease. + #math.equation(block: false, alt: "open parenthesis 2000 , 400 close parenthesis")[$( 2000 , 400 )$]; The largest annual increase in biomass, #math.equation(block: false, alt: "400")[$400$] tons, occurs when the biomass is #math.equation(block: false, alt: "2000")[$2000$] tons. + #figure(figph[parabola], alt: "parabola", caption: none) + #math.equation(block: false, alt: "4000 less than x less than or equal to 5000")[$4000 < x ≤ 5000$]; When there are too many fish, there will not be enough food to support all of them. The annual increase, #math.equation(block: false, alt: "I")[$I$], in the deer population in a national park depends on the size, #math.equation(block: false, alt: "x")[$x$], of the population that year, according to the function #math.equation(block: true, alt: "I equals f open parenthesis x close parenthesis equals 1.2 x minus 0.0002 x squared")[$I = f ( x ) = 1.2 x − 0.0002 x^(2)$] + Find the vertex of the graph. What does it tell us about the deer population? + Sketch the graph for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 7000")[$0 ≤ x ≤ 7000$]. + For what values of #math.equation(block: false, alt: "x")[$x$] does the deer population decrease rather than increase? Suggest a reason why the population might decrease. Many animals live in groups. A species of marmot found in Colorado lives in harems composed of a single adult male and several females with their young. The number of offspring each female can raise depends on the number of females in the harem. On average, if there are #math.equation(block: false, alt: "x")[$x$] females in the harem, each female can raise #math.equation(block: false, alt: "y equals 2 minus 0.4 x")[$y = 2 − 0.4 x$] young marmots each year. + Complete the table of values for the average number of offspring per female, and the total number of young marmots, #math.equation(block: false, alt: "A")[$A$], produced by the entire harem in one year.#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$" " " " 1 " " " "$]], [#math.equation(block: false, alt: "2")[$" " " " 2 " " " "$]], [#math.equation(block: false, alt: "3")[$" " " " 3 " " " "$]], [#math.equation(block: false, alt: "4")[$" " " " 4 " " " "$]], [#math.equation(block: false, alt: "5")[$" " " " 5 " " " "$]]), [#math.equation(block: false, alt: "y")[$y$]], [], [], [], [], [], [#math.equation(block: false, alt: "A")[$A$]], [], [], [], [], [], )) + Write a formula for #math.equation(block: false, alt: "A")[$A$] in terms of #math.equation(block: false, alt: "x")[$x$]. + Graph #math.equation(block: false, alt: "A")[$A$] as a function of #math.equation(block: false, alt: "x")[$x$]. + What is the maximum number of young marmots a harem can produce (on average)? What is the optimal number of female marmots per harem? + #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$" " " " 1 " " " "$]], [#math.equation(block: false, alt: "2")[$" " " " 2 " " " "$]], [#math.equation(block: false, alt: "3")[$" " " " 3 " " " "$]], [#math.equation(block: false, alt: "4")[$" " " " 4 " " " "$]], [#math.equation(block: false, alt: "5")[$" " " " 5 " " " "$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "1.6")[$1.6$]], [#math.equation(block: false, alt: "1.2")[$1.2$]], [#math.equation(block: false, alt: "0.8")[$0.8$]], [#math.equation(block: false, alt: "0.4")[$0.4$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "A")[$A$]], [#math.equation(block: false, alt: "1.6")[$1.6$]], [#math.equation(block: false, alt: "2.4")[$2.4$]], [#math.equation(block: false, alt: "2.4")[$2.4$]], [#math.equation(block: false, alt: "1.6")[$1.6$]], [#math.equation(block: false, alt: "0")[$0$]], )) + #math.equation(block: false, alt: "A equals x open parenthesis 2 minus 0.4 x close parenthesis")[$A = x ( 2 − 0.4 x )$] or #math.equation(block: false, alt: "A equals 2 x minus 0.4 x squared")[$A = 2 x − 0.4 x^(2)$] + #figure(figph[parabola], alt: "parabola", caption: none) + The maximum number of young marmots, on average, is #math.equation(block: false, alt: "2.5")[$2.5$]; the optimal number of female marmots is #math.equation(block: false, alt: "2.5")[$2.5$]. Greenshield's model for traffic flow assumes that the average speed, #math.equation(block: false, alt: "u")[$u$], of cars on a highway is a linear function of the traffic density, #math.equation(block: false, alt: "k")[$k$], in vehicles per mile, given by #math.equation(block: true, alt: "u equals u sub f open parenthesis 1 minus the fraction k over k sub j close parenthesis")[$u = u_(f) ( 1 − display(frac(k, k_(j))) )$] where #math.equation(block: false, alt: "u sub f")[$u_(f)$] is the free-flow speed and #math.equation(block: false, alt: "k sub j")[$k_(j)$] is the maximum density (the point when traffic jams). Then the traffic flow, #math.equation(block: false, alt: "q")[$q$], in vehicles per hour, is given by #math.equation(block: false, alt: "q equals u k")[$q = u k$]. + Write a formula for #math.equation(block: false, alt: "q")[$q$] as a function of #math.equation(block: false, alt: "k")[$k$]. + If the free-flow speed is #math.equation(block: false, alt: "70")[$70$] mph and the maximum density is #math.equation(block: false, alt: "240")[$240$] vehicles per mile, graph #math.equation(block: false, alt: "q")[$q$] as a function of #math.equation(block: false, alt: "k")[$k$]. + What value of #math.equation(block: false, alt: "k")[$k$] gives the maximum traffic flow? What is the average speed of vehicles at that density? After touchdown, the distance the space shuttle travels is given by #math.equation(block: true, alt: "d equals v T plus the fraction v squared over 2 a")[$d = v T + frac(v^(2), 2 a)$] where #math.equation(block: false, alt: "v")[$v$] is the shuttle's velocity in ft/sec at touchdown, #math.equation(block: false, alt: "T")[$T$] is the pilot's reaction time before the brakes are applied, and #math.equation(block: false, alt: "a")[$a$] is the shuttle's deceleration. + Graph #math.equation(block: false, alt: "d equals f open parenthesis v close parenthesis")[$d = f ( v )$] for #math.equation(block: false, alt: "T equals 0.5")[$T = 0.5$] seconds and #math.equation(block: false, alt: "a equals 12 ft/sec squared")[$a = 12 attach(" ft/sec", t: 2)$]. Find the coordinates of the vertex and the horizontal intercepts. Explain their meaning, if any, in this context. + The runway at Edwards Air Force base is #math.equation(block: false, alt: "15 , 000")[$15 , 000$] feet long. What is the maximum velocity the shuttle can have at touchdown and still stop on the runway? + #figure(figph[parabola], alt: "parabola", caption: none) #linebreak() Vertex: #math.equation(block: false, alt: "open parenthesis minus 6 , minus 1.5 close parenthesis")[$( − 6 , − 1.5 )$]; Horizontal intercepts #math.equation(block: false, alt: "open parenthesis minus 12 , 0 close parenthesis")[$( − 12 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]. The point #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] means that no distance is required to stop a plane that is not moving. + #math.equation(block: false, alt: "594")[$594$] ft/sec When setting the pump pressure at the engine, firefighters must take into account the pressure loss due to friction inside the fire hose. For every #math.equation(block: false, alt: "100")[$100$] feet of hoseline, a hose of diameter #math.equation(block: false, alt: "2.5")[$2.5$] inches loses pressure according to the formula #math.equation(block: true, alt: "L equals open brace 2 Q squared plus Q ,, Q greater than or equal to 1; 2 Q squared plus the fraction 1 over 2 Q ,, Q less than 1")[$L = \{ 2 Q^(2) + Q , & Q ≥ 1 \ 2 Q^(2) + display(frac(1, 2)) Q , & Q < 1$] where #math.equation(block: false, alt: "Q")[$Q$] is the water flow in hundreds of gallos per minute. The friction loss, #math.equation(block: false, alt: "L")[$L$], is measured in pounds per square inch (psi). (Source: www.hcc.hawaii.edu/~jkemmer) + Graph #math.equation(block: false, alt: "L equals g open parenthesis Q close parenthesis")[$L = g ( Q )$] on the domain #math.equation(block: false, alt: "open bracket 0 , 5 close bracket")[$[ 0 , 5 ]$]. + The firefighters have unrolled #math.equation(block: false, alt: "600")[$600$] feet of #math.equation(block: false, alt: "2.5")[$2.5$]-inch-diameter hose, and they would like to deliver water at a rate of #math.equation(block: false, alt: "200")[$200$] gallons per minute, with nozzle pressure at #math.equation(block: false, alt: "100")[$100$] psi. They must add the friction loss to the nozzle pressure to calculate the engine pressure required. What should the engine pressure be? For Problems 15–16, find the coordinates of the vertex. Decide whether the vertex is a maximum point or a minimum point on the graph. + #math.equation(block: false, alt: "y equals 2 plus 3 x minus x squared")[$y = 2 + 3 x − x^(2)$] + #math.equation(block: false, alt: "y equals the fraction 1 over 2 x squared minus the fraction 2 over 3 x plus the fraction 1 over 3")[$y = display(frac(1, 2)) x^(2) − display(frac(2, 3)) x + display(frac(1, 3))$] + #math.equation(block: false, alt: "y equals 2.3 minus 7.2 x minus 0.8 x squared")[$y = 2.3 − 7.2 x − 0.8 x^(2)$] + #math.equation(block: false, alt: "open parenthesis the fraction 3 over 2 , the fraction 17 over 4 close parenthesis")[$( display(frac(3, 2)) , display(frac(17, 4)) )$], maximum + #math.equation(block: false, alt: "open parenthesis the fraction 2 over 3 , the fraction 1 over 9 close parenthesis")[$( display(frac(2, 3)) , display(frac(1, 9)) )$], minimum + #math.equation(block: false, alt: "open parenthesis minus 4.5 , 18.5 close parenthesis")[$( − 4.5 , 18.5 )$], maximum + #math.equation(block: false, alt: "y equals 3 minus 5 x plus x squared")[$y = 3 − 5 x + x^(2)$] + #math.equation(block: false, alt: "y equals the fraction minus 3 over 4 x squared plus the fraction 1 over 2 x minus the fraction 1 over 4")[$y = display(frac(− 3, 4)) x^(2) + display(frac(1, 2)) x − display(frac(1, 4))$] + #math.equation(block: false, alt: "y equals 5.1 minus 0.2 x plus 4.6 x squared")[$y = 5.1 − 0.2 x + 4.6 x^(2)$] In Problems 17–26, + Find the coordinates of the intercepts and the vertex. + Sketch the graph by hand. + Use your calculator to verify your graph. #math.equation(block: true, alt: "y equals minus 2 x squared plus 7 x plus 4")[$y = − 2 x^(2) + 7 x + 4$] + #math.equation(block: false, alt: "x")[$x$]-intercepts: #math.equation(block: false, alt: "open parenthesis the fraction minus 1 over 2 , 0 close parenthesis")[$( frac(− 1, 2) , 0 )$] and #math.equation(block: false, alt: "open parenthesis 4 , 0 close parenthesis")[$( 4 , 0 )$]; #math.equation(block: false, alt: "y")[$y$]-intercept: #math.equation(block: false, alt: "open parenthesis 0 , 4 close parenthesis")[$( 0 , 4 )$]; vertex: #math.equation(block: false, alt: "open parenthesis the fraction 7 over 4 , the fraction 81 over 8 close parenthesis")[$( frac(7, 4) , frac(81, 8) )$] + #figure(figph[parabola], alt: "parabola", caption: none) #math.equation(block: true, alt: "y equals minus 3 x squared plus 2 x plus 8")[$y = − 3 x^(2) + 2 x + 8$] #math.equation(block: true, alt: "y equals 0.6 x squared plus 0.6 x minus 1.2")[$y = 0.6 x^(2) + 0.6 x − 1.2$] + #math.equation(block: false, alt: "x")[$x$]-intercepts: #math.equation(block: false, alt: "open parenthesis minus 2 , 0 close parenthesis")[$( − 2 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]; #math.equation(block: false, alt: "y")[$y$]-intercept: #math.equation(block: false, alt: "open parenthesis 0 , minus 1.2 close parenthesis")[$( 0 , − 1.2 )$]; vertex: #math.equation(block: false, alt: "open parenthesis minus 0.5 , minus 1.35 close parenthesis")[$( − 0.5 , − 1.35 )$] + #figure(figph[parabola], alt: "parabola", caption: none) #math.equation(block: true, alt: "y equals 0.5 x squared minus 0.25 x minus 0.75")[$y = 0.5 x^(2) − 0.25 x − 0.75$] #math.equation(block: true, alt: "y equals x squared plus 4 x plus 7")[$y = x^(2) + 4 x + 7$] + No #math.equation(block: false, alt: "x")[$x$]-intercepts; #math.equation(block: false, alt: "y")[$y$]-intercept: #math.equation(block: false, alt: "open parenthesis 0 , 7 close parenthesis")[$( 0 , 7 )$]; vertex: #math.equation(block: false, alt: "open parenthesis minus 2 , 3 close parenthesis")[$( − 2 , 3 )$] + #figure(figph[parabola], alt: "parabola", caption: none) #math.equation(block: true, alt: "y equals x squared minus 6 x plus 10")[$y = x^(2) − 6 x + 10$] #math.equation(block: true, alt: "y equals x squared plus 2 x minus 1")[$y = x^(2) + 2 x − 1$] + #math.equation(block: false, alt: "x")[$x$]-intercepts: #math.equation(block: false, alt: "open parenthesis minus 1 plus or minus the square root of 2 , 0 close parenthesis")[$( − 1 ± sqrt(2) , 0 )$]; #math.equation(block: false, alt: "y")[$y$]-intercept: #math.equation(block: false, alt: "open parenthesis 0 , minus 1 close parenthesis")[$( 0 , − 1 )$]; vertex: #math.equation(block: false, alt: "open parenthesis minus 1 , minus 2 close parenthesis")[$( − 1 , − 2 )$] + #figure(figph[parabola], alt: "parabola", caption: none) #math.equation(block: true, alt: "y equals x squared minus 6 x plus 2")[$y = x^(2) − 6 x + 2$] #math.equation(block: true, alt: "y equals minus 2 x squared plus 6 x minus 3")[$y = − 2 x^(2) + 6 x − 3$] + #math.equation(block: false, alt: "x")[$x$]-intercepts: #math.equation(block: false, alt: "open parenthesis the fraction 3 plus or minus the square root of 3 over 2 , 0 close parenthesis")[$( display(frac(3 ± sqrt(3), 2)) , 0 )$]; #math.equation(block: false, alt: "y")[$y$]-intercept: #math.equation(block: false, alt: "open parenthesis 0 , minus 3 close parenthesis")[$( 0 , − 3 )$]; vertex: #math.equation(block: false, alt: "open parenthesis the fraction 3 over 2 , the fraction 3 over 2 close parenthesis")[$( display(frac(3, 2)) , display(frac(3, 2)) )$] + #figure(figph[parabola], alt: "parabola", caption: none) #math.equation(block: true, alt: "y equals minus 2 x squared minus 8 x minus 5")[$y = − 2 x^(2) − 8 x − 5$] + Graph the three functions #math.equation(block: true, alt: "f open parenthesis x close parenthesis, equals x squared minus 6 x plus 5; g open parenthesis x close parenthesis, equals x squared minus 6 x plus 9; h open parenthesis x close parenthesis, equals x squared minus 6 x plus 12")[$f ( x ) & = x^(2) − 6 x + 5 \ g ( x ) & = x^(2) − 6 x + 9 \ h ( x ) & = x^(2) − 6 x + 12$] in the window #math.equation(block: true, alt: "Xmin, equals minus 2, Xmax equals 7.4; Ymin, equals minus 5, Ymax equals 15")[$"Xmin" & = − 2 & & "Xmax" = 7.4 \ "Ymin" & = − 5 & & "Ymax" = 15$] Use the #strong[Trace] to locate the #math.equation(block: false, alt: "x")[$x$]-intercepts of each graph. + Set #math.equation(block: false, alt: "y equals 0")[$y = 0$] for each of the equations in part (a) and calculate the discriminant. What does the discriminant tell you about the solutions of the equation? How does your answer relate to the graphs in part (a)? + #figure(figph[three parabolas], alt: "three parabolas", caption: none) #linebreak() #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 6 x plus 5")[$f ( x ) = x^(2) − 6 x + 5$]: #math.equation(block: false, alt: "x")[$x$]-intercepts #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 5 , 0 close parenthesis")[$( 5 , 0 )$]; #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals x squared minus 6 x plus 9")[$g ( x ) = x^(2) − 6 x + 9$]: #math.equation(block: false, alt: "x")[$x$]-intercept #math.equation(block: false, alt: "open parenthesis 3 , 0 close parenthesis")[$( 3 , 0 )$]; #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals x squared minus 6 x plus 12")[$h ( x ) = x^(2) − 6 x + 12$]: No #math.equation(block: false, alt: "x")[$x$]-intercept. + #math.equation(block: false, alt: "16 , 0 , minus 12")[$16 , 0 , − 12$]: #math.equation(block: false, alt: "D equals 16")[$D = 16$] means that there are two rational #math.equation(block: false, alt: "x")[$x$]-intercepts, #math.equation(block: false, alt: "D equals 0")[$D = 0$] means that there is exactly one #math.equation(block: false, alt: "x")[$x$]-intercept, #math.equation(block: false, alt: "D equals minus 12")[$D = − 12$] means that there is no #math.equation(block: false, alt: "x")[$x$]-intercept. + Graph the three functions #math.equation(block: true, alt: "F open parenthesis x close parenthesis, equals 3 minus 2 x minus x squared; G open parenthesis x close parenthesis, equals minus 1 minus 2 x minus x squared; H open parenthesis x close parenthesis, equals minus 4 minus 2 x minus x squared")[$F ( x ) & = 3 − 2 x − x^(2) \ G ( x ) & = − 1 − 2 x − x^(2) \ H ( x ) & = − 4 − 2 x − x^(2)$] in the window #math.equation(block: true, alt: "Xmin, equals minus 6.4, Xmax equals 3; Ymin, equals minus 10, Ymax equals 5")[$"Xmin" & = − 6.4 & & "Xmax" = 3 \ "Ymin" & = − 10 & & "Ymax" = 5$] Use the #strong[Trace] to locate the #math.equation(block: false, alt: "x")[$x$]-intercepts of each graph. + Set #math.equation(block: false, alt: "y equals 0")[$y = 0$] for each of the equations in part (a) and calculate the discriminant. What does the discriminant tell you about the solutions of the equation? How does your answer relate to the graphs in part (a)? For Problems 29–34, use the discriminant to determine the nature of the solutions of each equation. #math.equation(block: true, alt: "3 x squared plus 26 equals 17 x")[$3 x^(2) + 26 = 17 x$] Two complex solutions #math.equation(block: true, alt: "4 x squared plus 23 x equals 19")[$4 x^(2) + 23 x = 19$] #math.equation(block: true, alt: "16 x squared minus 712 x plus 7921 equals 0")[$16 x^(2) − 712 x + 7921 = 0$] One repeated rational solution #math.equation(block: true, alt: "121 x squared plus 1254 x plus 3249 equals 0")[$121 x^(2) + 1254 x + 3249 = 0$] #math.equation(block: true, alt: "65.2 x equals 13.2 x squared plus 41.7")[$65.2 x = 13.2 x^(2) + 41.7$] Two distinct real solutions #math.equation(block: true, alt: "0.03 x squared equals 0.05 x minus 0.12")[$0.03 x^(2) = 0.05 x − 0.12$] For problems 35–38, use the discriminant to decide if we can solve the equation by factoring. #math.equation(block: true, alt: "3 x squared minus 7 x plus 6 equals 0")[$3 x^(2) − 7 x + 6 = 0$] No #math.equation(block: true, alt: "6 x squared minus 11 x minus 7 equals 0")[$6 x^(2) − 11 x − 7 = 0$] #math.equation(block: true, alt: "15 x squared minus 52 x minus 32 equals 0")[$15 x^(2) − 52 x − 32 = 0$] Yes #math.equation(block: true, alt: "17 x squared plus 65 x minus 12 equals 0")[$17 x^(2) + 65 x − 12 = 0$] For Problems 39–42, + Given one solution of a quadratic equation with rational coefficients, find the other solution. + Write a quadratic equation that has those solutions. #math.equation(block: true, alt: "2 plus the square root of 5")[$2 + sqrt(5)$] + #math.equation(block: false, alt: "2 minus the square root of 5")[$2 − sqrt(5)$] + #math.equation(block: false, alt: "x squared minus 4 x minus 1 equals 0")[$x^(2) − 4 x − 1 = 0$] #math.equation(block: true, alt: "3 minus the square root of 2")[$3 − sqrt(2)$] #math.equation(block: true, alt: "4 minus 3 the square root of 2")[$4 − 3 sqrt(2)$] + #math.equation(block: false, alt: "4 plus 3 the square root of 2")[$4 + 3 sqrt(2)$] + #math.equation(block: false, alt: "x squared minus 8 x minus 2 equals 0")[$x^(2) − 8 x − 2 = 0$] #math.equation(block: true, alt: "5 plus 2 the square root of 3")[$5 + 2 sqrt(3)$] For Problems 43 and 44, match each equation with one of the eight graphs shown. #figure(figph[eight parabolas], alt: "eight parabolas", caption: none) + #math.equation(block: false, alt: "y equals 1 minus x squared")[$y = 1 − x^(2)$] + #math.equation(block: false, alt: "y equals open parenthesis x plus 2 close parenthesis squared")[$y = ( x + 2 )^(2)$] + #math.equation(block: false, alt: "y equals 2 x squared")[$y = 2 x^(2)$] + #math.equation(block: false, alt: "y equals open parenthesis x minus 4 close parenthesis open parenthesis x plus 2 close parenthesis")[$y = ( x − 4 ) ( x + 2 )$] + IV + V + I + VII + #math.equation(block: false, alt: "y equals minus 2 minus open parenthesis x minus 2 close parenthesis squared")[$y = − 2 − ( x − 2 )^(2)$] + #math.equation(block: false, alt: "y equals x minus x squared")[$y = x − x^(2)$] + #math.equation(block: false, alt: "y equals x squared minus 4")[$y = x^(2) − 4$] + #math.equation(block: false, alt: "y equals minus 0.5 x squared")[$y = − 0.5 x^(2)$] + Write an equation for a parabola that has #math.equation(block: false, alt: "x")[$x$]-intercepts at #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$] and #math.equation(block: false, alt: "open parenthesis minus 3 , 0 close parenthesis")[$( − 3 , 0 )$]. What is the equation of the parabola's axis of symmetry? + Write an equation for another parabola that has the same #math.equation(block: false, alt: "x")[$x$]-intercepts. What is the equation of the parabola's axis of symmetry? + #math.equation(block: false, alt: "y equals x squared plus x minus 6")[$y = x^(2) + x − 6$]; #math.equation(block: false, alt: "x equals the fraction minus 1 over 2")[$" " x = display(frac(− 1, 2))$] + #math.equation(block: false, alt: "y equals 2 x squared plus 2 x minus 12")[$y = 2 x^(2) + 2 x − 12$]; #math.equation(block: false, alt: "x equals the fraction minus 1 over 2")[$" " x = display(frac(− 1, 2))$] + Write an equation for a parabola that opens upward and has #math.equation(block: false, alt: "x")[$x$]-intercepts at #math.equation(block: false, alt: "open parenthesis minus 1 , 0 close parenthesis")[$( − 1 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 4 , 0 close parenthesis")[$( 4 , 0 )$]. What is the equation of the parabola's axis of symmetry? + Write an equation for a parabola that opens downward and has #math.equation(block: false, alt: "x")[$x$]-intercepts #math.equation(block: false, alt: "open parenthesis minus 1 , 0 close parenthesis")[$( − 1 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 4 , 0 close parenthesis")[$( 4 , 0 )$]. What is the equation of its axis of symmetry? + Graph the functions in the same window on your calculator: #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x squared plus 2 x , g open parenthesis x close parenthesis equals x squared plus 4 x ,; h open parenthesis x close parenthesis equals x squared plus 6 x , j open parenthesis x close parenthesis equals x squared plus 8 x")[$f ( x ) = x^(2) + 2 x , " " " " " " g ( x ) = x^(2) + 4 x , \ h ( x ) = x^(2) + 6 x , " " " " " " j ( x ) = x^(2) + 8 x$] + Find the vertex of each graph in part (a) and plot the points. + Find the equation of the curve in part (b). + Show that the vertex of #math.equation(block: false, alt: "y equals x squared plus 2 k x")[$y = x^(2) + 2 k x$] lies on the curve for any value of #math.equation(block: false, alt: "k")[$k$]. + #figure(figph[four parabolas], alt: "four parabolas", caption: none) + #figure(figph[four points], alt: "four points", caption: none) #linebreak() #math.equation(block: false, alt: "open parenthesis minus 1 , minus 1 close parenthesis , open parenthesis minus 2 , minus 4 close parenthesis , open parenthesis minus 3 , minus 9 close parenthesis , open parenthesis minus 4 , minus 16 close parenthesis")[$( − 1 , − 1 ) , ( − 2 , − 4 ) , ( − 3 , − 9 ) , ( − 4 , − 16 )$] + #math.equation(block: false, alt: "y equals minus x squared")[$y = − x^(2)$] + The vertex of #math.equation(block: false, alt: "y equals x squared plus 2 k x")[$y = x^(2) + 2 k x$] is #math.equation(block: false, alt: "open parenthesis minus k , minus k squared close parenthesis")[$( − k , − k^(2) )$] + Graph the functions in the same window on your calculator: #math.equation(block: true, alt: "F open parenthesis x close parenthesis equals x minus the fraction 1 over 2 x squared , G open parenthesis x close parenthesis equals 3 x minus the fraction 1 over 2 x squared ,; H open parenthesis x close parenthesis equals 5 x minus the fraction 1 over 2 x squared , J open parenthesis x close parenthesis equals 7 x minus the fraction 1 over 2 x squared")[$F ( x ) = x − frac(1, 2) x^(2) , " " " " " " G ( x ) = 3 x − frac(1, 2) x^(2) , \ H ( x ) = 5 x − frac(1, 2) x^(2) , " " " " " " J ( x ) = 7 x − frac(1, 2) x^(2)$] + Find the vertex of each graph in part (a) and plot the points. + Find the equation of the curve in part (b). + Show that the vertex of #math.equation(block: false, alt: "y equals k x minus the fraction 1 over 2 x squared")[$y = k x − display(frac(1, 2)) x^(2)$] lies on the curve for any value of #math.equation(block: false, alt: "k")[$k$]. Because of air resistance, the path of a kicked soccer ball is not actually parabolic. However, both the horizontal and vertical coordinates of points on its trajectory can be approximated by quadratic functions. For a soccer ball kicked from the ground, these functions are #math.equation(block: true, alt: "x equals f open parenthesis t close parenthesis equals 12.8 t minus 1.3 t squared; y equals g open parenthesis t close parenthesis equals 17.28 t minus 4.8 t squared")[$x = f ( t ) = 12.8 t − 1.3 t^(2) \ y = g ( t ) = 17.28 t − 4.8 t^(2)$] where #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] are given in meters and #math.equation(block: false, alt: "t")[$t$] is the number of seconds since the ball was kicked. + Fill in the table.#figure(table( columns: 9, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1.0")[$1.0$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "2.0")[$2.0$]], [#math.equation(block: false, alt: "2.5")[$2.5$]], [#math.equation(block: false, alt: "3.0")[$3.0$]], [#math.equation(block: false, alt: "3.5")[$3.5$]]), [#math.equation(block: false, alt: "x")[$x$]], [], [], [], [], [], [], [], [], [#math.equation(block: false, alt: "y")[$y$]], [], [], [], [], [], [], [], [], )) + Plot the points #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] from your table and connect them with a smooth curve to represent the path of the ball. + Use your graph to estimate the maximum height of the ball. + Estimate the horizontal distance traveled by the ball before it strikes the ground. + Using the formula given for #math.equation(block: false, alt: "y")[$y$], determine how long the ball is in the air. + Use your answer from part (e) and the formula for #math.equation(block: false, alt: "x")[$x$] to find the horizontal distance traveled by the ball before it strikes the ground. + Use the formula given for #math.equation(block: false, alt: "y")[$y$] to find the maximum height for the ball. + #figure(table( columns: 9, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1.0")[$1.0$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "2.0")[$2.0$]], [#math.equation(block: false, alt: "2.5")[$2.5$]], [#math.equation(block: false, alt: "3.0")[$3.0$]], [#math.equation(block: false, alt: "3.5")[$3.5$]]), [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "6.075")[$6.075$]], [#math.equation(block: false, alt: "11.5")[$11.5$]], [#math.equation(block: false, alt: "16.275")[$16.275$]], [#math.equation(block: false, alt: "20.4")[$20.4$]], [#math.equation(block: false, alt: "23.875")[$23.875$]], [#math.equation(block: false, alt: "26.7")[$26.7$]], [#math.equation(block: false, alt: "28.875")[$28.875$]], [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "7.44")[$7.44$]], [#math.equation(block: false, alt: "12.48")[$12.48$]], [#math.equation(block: false, alt: "15.12")[$15.12$]], [#math.equation(block: false, alt: "15.36")[$15.36$]], [#math.equation(block: false, alt: "13.2")[$13.2$]], [#math.equation(block: false, alt: "8.64")[$8.64$]], [#math.equation(block: false, alt: "1.68")[$1.68$]], )) + #figure(figph[soccer ball path], alt: "soccer ball path", caption: none) + #math.equation(block: false, alt: "y approximately equals 15.4")[$y ≈ 15.4$] m + #math.equation(block: false, alt: "x approximately equals 30")[$x ≈ 30$] m + #math.equation(block: false, alt: "3.6")[$3.6$] sec + #math.equation(block: false, alt: "x approximately equals 29.2")[$x ≈ 29.2$] m + #math.equation(block: false, alt: "y approximately equals 15.55")[$y ≈ 15.55$] m How far can you throw a baseball? The distance depends on the initial speed of the ball, #math.equation(block: false, alt: "v")[$v$], and on the angle at which you throw it. For maximum range, you should throw the ball at #math.equation(block: false, alt: "45 degrees")[$45^(∘)$]. + If there were no air resistance, the height, #math.equation(block: false, alt: "x")[$x$], of the ball t seconds after its release would be given in meters by the function #math.equation(block: true, alt: "h equals f open parenthesis t close parenthesis equals the fraction v t over the square root of 2 minus the fraction g t squared over 2")[$h = f ( t ) = frac(v t, sqrt(2)) − frac(g t^(2), 2)$] where #math.equation(block: false, alt: "g")[$g$] is the acceleration due to gravity. Find an expression for the total time the ball is in the air. (#emph[Hint]: Set #math.equation(block: false, alt: "h equals 0")[$h = 0$] and solve for #math.equation(block: false, alt: "t")[$t$] in terms of the other variables.) + At time #math.equation(block: false, alt: "t")[$t$], the ball has traveled a horizontal distance #math.equation(block: false, alt: "d")[$d$] given by #math.equation(block: true, alt: "d equals the fraction v t over the square root of 2")[$d = frac(v t, sqrt(2))$] Find an expression for the range of the ball in terms of its velocity, #math.equation(block: false, alt: "v")[$v$]. (#emph[Hint]: In part (a), you found an expression for #math.equation(block: false, alt: "t")[$t$] when #math.equation(block: false, alt: "h equals 0")[$h = 0$]. Use that value of #math.equation(block: false, alt: "t")[$t$] to calculate #math.equation(block: false, alt: "d")[$d$] when #math.equation(block: false, alt: "h equals 0")[$h = 0$].) + The fastest baseball pitch on record was #math.equation(block: false, alt: "45")[$45$] meters per second, or about #math.equation(block: false, alt: "100")[$100$] miles per hour. Use your formula from part (b) to calculate the theoretical range of such a pitch. The value of #math.equation(block: false, alt: "g")[$g$] is #math.equation(block: false, alt: "9.8")[$9.8$]. + The maximum distance a baseball has actually been thrown is #math.equation(block: false, alt: "136")[$136$] meters. Can you explain the discrepancy between this figure and your answer to part (c)?