4.1 Root and Radical Functions

In Sections 1.2, 1.3 and 1.4, we studied constant, linear, absolute value,¹ and quadratic functions. Constant, linear and quadratic functions were specific examples of polynomial functions, which we studied in generality in Chapter 2. Chapter 2 culminated with the Real Factorization Theorem, Theorem 2.18, which says that all polynomial functions with real coefficients can be thought of as products of linear and quadratic functions. Our next step was to enlarge our field² of study to rational functions in Chapter 3. Being quotients of polynomials, we can ultimately view this family of functions as being built up of linear and quadratic functions as well. So in some sense, Sections 1.2, 1.3 and 1.4 along with Chapters 2 and 3 can be thought of as an exhaustive study of linear and quadratic³ functions. We now turn our attention to functions involving radicals which cannot be written in terms of linear functions. For a more detailed review of the basics of roots and radicals, we refer the reader to Sections A.2 and A.13.

Root Functions

As with polynomial functions and rational functions, we begin our study of functions involving radical with a special family of functions: the (principal) root functions.

Definition

Let $n ∈ ℕ$ with $n greater than or equal to 2$ . The $n$ th (principal) root function is the function $f (x) = \sqrt[n]{x}$ .

NOTE: If $n$ is even, the domain of $f$ is $open bracket 0, ∞ close parenthesis$ ; if $n$ is odd, the domain of $f$ is $open parenthesis minus ∞, ∞ close parenthesis$ .

The domain restriction for even indexed roots means that, once again, we are restricting our attention to real numbers.⁴ We graph a few members of the root function family below, and quickly notice that, as with the monomial, and, more generally, the Laurent monomial functions, the behavior of the root functions depends primarily on whether the root is even or odd.

In addition to having the common domain of $open bracket 0, ∞ close parenthesis$ , the graphs of $f (x) = \sqrt[n]{x}$ for even indices $n$ all share the points $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 1, 1 close parenthesis$ . As $n$ increases, the functions become `steeper' near the $y$ -axis and `flatter' as $x → ∞$ . To show $f ⁢ open parenthesis x close parenthesis → ∞$ as $x → ∞$ , we show, more generally, the range of $f$ is $open bracket 0, ∞ close parenthesis$ . Indeed, if $c greater than or equal to 0$ is a real number, then $f (c^{n}) = \sqrt[n]{c^{n}} = c$ so $c$ is in the range of $f$ . Note that $f$ is increasing: that is, if $a < b$ , then $f (a) = \sqrt[n]{a} < \sqrt[n]{b} = f (b)$ . This property is useful in solving certain types of polynomial inequalities.⁵

Coordinate-plane figure. — **Figure 4.1** $y = \sqrt{x}$

The functions $f (x) = \sqrt[n]{x}$ for odd natural numbers $n greater than or equal to 3$ also follow a predictable trend - steepening near $x equals 0$ and flattening as $x → minus ∞$ and $x → minus ∞$ . The range for these functions is $open parenthesis minus ∞, ∞ close parenthesis$ since if $c$ is any real number, $f (c^{n}) = \sqrt[n]{c^{n}} = c$ , so $c$ is in the range of $f$ . Like the even indexed roots, the odd indexed roots are also increasing. Moreover, these graphs appear to be symmetric about the origin. Sure enough, when $n$ is odd, $f (- x) = \sqrt[n]{- x} = - \sqrt[n]{x} = - f (x)$ so $f$ is an odd function.

At this point, you're probably expecting a theorem like Theorems 1.2, 1.3, 2.1, 3.1 - that is, a theorem which tells us how to obtain the graph of $F (x) = a \sqrt[n]{x - h} + k$ from the graph of $f (x) = \sqrt[n]{x}$ - and you would not be wrong. Here, however, we need to add an extra parameter ` $b$ ' to the recipe and discuss functions of the form $F (x) = a \sqrt[n]{b x - h} + k$ . The reason is that, with all of the previous function families, we were always able to factor out the coefficient of $x$ . We list some examples of this below, and invite the reader to revisit other examples in the text:

$F (x) = | 6 - 2 x | = | - 2 x + 6 | = | - 2 (x + 3) | = | - 2 | | x + 3 | = 2 | x + 3 |$ .
$F (x) = {(2 x - 1)}^{2} + 1 = {[2 (x - \frac{1}{2})]}^{2} + 1 = {(2)}^{2} {(x - \frac{1}{2})}^{2} + 1 = 4 {(x - \frac{1}{2})}^{2} + 1$
$F (x) = \frac{2}{{(1 - x)}^{3}} - 5 = \frac{2}{{[(- 1) (x - 1)]}^{3}} - 5 = \frac{2}{{(- 1)}^{3} {(x - 1)}^{3}} - 5 = \frac{2}{- {(x - 1)}^{3}} - 5 = \frac{- 2}{{(x - 1)}^{3}} - 5$ .

For a function like $F (x) = \sqrt{4 x - 12} + 1 = \sqrt{4 (x - 6)} + 1 = \sqrt{4} \sqrt{x - 3} + 1 = 2 \sqrt{x - 3} + 1$ , this approach works fine. However, if the coefficient of $x$ is negative, for example, $F (x) = \sqrt{1 - x} = \sqrt{(- 1) (x - 1)}$ we get stuck the product rule for radicals doesn't extend to negative quantities when the index is even.⁶ Hence we add an extra parameter which means we have an extra step. We state Theorem 4.1 below.

Theorem

For real numbers $a$ , $b$ , $h$ , and $k$ with $a, b not equal to 0$ , the graph of $F (x) = a \sqrt[n]{b x - h} + k$ can be obtained from the graph of $f (x) = \sqrt[n]{x}$ by performing the following operations, in sequence:

add $h$ to each of the $x$ -coordinates of the points on the graph of $f$ . This results in a horizontal shift to the right if $h greater than 0$ or left if $h less than 0$ .
NOTE: This transforms the graph of $y = \sqrt[n]{x}$ to $y = \sqrt[n]{x - h}$ .
divide the $x$ -coordinates of the points on the graph obtained in Step 1 by $b$ . This results in a horizontal scaling, but may also include a reflection about the $y$ -axis if $b less than 0$ .
NOTE: This transforms the graph of $y = \sqrt[n]{x - h}$ to $y = \sqrt[n]{b x - h}$ .
multiply the $y$ -coordinates of the points on the graph obtained in Step 2 by $a$ . This results in a vertical scaling, but may also include a reflection about the $x$ -axis if $a less than 0$ .
NOTE: This transforms the graph of $y = \sqrt[n]{b x - h}$ to $y = a \sqrt[n]{b x - h}$ .
add $k$ to each of the $y$ -coordinates of the points on the graph obtained in Step 3. This results in a vertical shift up if $k greater than 0$ or down if $k less than 0$ .
NOTE: This transforms the graph of $y = a \sqrt[n]{b x - h}$ to $y = a \sqrt[n]{b x - h} + k$ .

Proof. As usual, we `build' the graph of $F (x) = a \sqrt[n]{b x - h} + k$ starting with the graph of $f (x) = \sqrt[n]{x}$ one step at a time. First, we consider the graph of $F_{1} (x) = \sqrt{x - h}$ . A generic point on the graph of $F sub 1$ looks like $(x, \sqrt[n]{x - h})$ . Note that if $n$ is odd, $x$ can be any real number whereas if $n$ is even $x minus h greater than or equal to 0$ so $x \geq h$ . If we let $c = x - h$ , then $x = c + h$ and we can change (dummy) variables⁷ and obtain a new representation of the point: $(c + h, \sqrt[n]{c})$ . Note that if $n$ is odd, $x$ and $c$ vary through all real numbers; if $n$ is even, $x \geq h$ and, hence, $c greater than or equal to 0$ . Since a generic point on the graph of $f (x) = \sqrt[n]{x}$ can be represented as $(c, \sqrt[n]{c})$ for applicable values of $c$ , we see that we can obtain every point on the graph of $F sub 1$ by adding $h$ to each $x$ -coordinate of the graph of $f$ , establishing step 1 of the theorem.

Proceeding to (the new!) step 2, a point on the graph of $F_{2} (x) = \sqrt[n]{b x - h}$ has the form $(x, \sqrt[n]{b x - h})$ . If $n$ is odd, as usual, $x$ can vary through all real numbers. If $n$ is even, we require $b ⁢ x minus h greater than or equal to 0$ or $b x \geq h$ . If $b greater than 0$ , this gives $x \geq \frac{h}{b}$ . If, on the other hand, $b less than 0$ , the we have $x \leq \frac{h}{b}$ . Let $c = b x$ and since by assumption $b not equal to 0$ , we have $x = \frac{c}{b}$ . Once again, we change dummy variables from $x$ to $c$ and describe a generic point on the graph of $F sub 2$ as $(\frac{c}{b}, \sqrt[n]{c - h})$ . If $n$ is odd, $x$ and $c$ can vary through all real numbers. If $n$ is even and $b greater than 0$ , then $x \geq \frac{h}{b}$ and, hence, $c = b x \geq h$ ; if $b less than 0$ , then $x \leq \frac{h}{b}$ also gives $c = b x \geq h$ . Since a generic point on the graph of $F sub 1$ can be represented as $(c, \sqrt{c - h})$ for applicable values of $c$ , we see we can obtain every point on the graph of $F sub 2$ by dividing every $x$ -coordinate on the graph of $F sub 1$ by $b$ , as per step 2 of the theorem.

The proof of steps 3 and 4 of Theorem 4.1 are identical to the proof of Theorem 2.1 (just with $\sqrt[n]{\cdot}$ instead of ${(\cdot)}^{n}$ ) so we invite the reader to work through the details on their own.

We demonstrate Theorem 4.1 in the following example.

Example 1

Theorem 4.1 to graph the following functions. Label at least three points on the graph. State the domain and range using interval notation.

$f (x) = 1 - 2 \sqrt[3]{x + 3}$
$g (t) = \frac{\sqrt{1 - 2 t}}{4}$

We begin by rewriting the expression for $f (x)$ in the form prescribed Theorem 4.1: $f (x) = - 2 \sqrt[3]{x + 3} + 1$ . We identify $n equals 3$ , $a = - 2$ , $b equals 1$ , $h = - 3$ and $k equals 1$ .
Step 1: add $minus 3$ to each of the $x$ -coordinates of each of the points on the graph of $y = \sqrt[3]{x}$ :
Figure 4.7 $y = \sqrt[3]{x}$
$→ add minus 3 to each x -coordinate$
Figure 4.8 $y = \sqrt[3]{x + 3}$
$open parenthesis minus 1, 1 close parenthesis, open parenthesis 0, 0 close parenthesis, open parenthesis 1, 1 close parenthesis$
$(- 4, - 1), (- 3, 0), (- 2, 1)$
Since $b equals 1$ , we can proceed to Step 3 (since dividing a real by $1$ just results in the same real number.)
Step 3: multiply each of the $y$ -coordinates of each point on the graph of $y = \sqrt[3]{x + 3}$ by $minus 2$ :
Figure 4.9 $y = \sqrt[3]{x + 3}$
$\overset{multiply each y -value by - 2}{\overset{}{\to}}$
Figure 4.10 $y = - 2 \sqrt[3]{x + 3}$
$(- 4, - 1), (- 3, 0), (- 2, 1)$
$(- 4, 2), (- 3, 0), (- 2, - 2)$
Step 4: add $1$ to $y$ -coordinates of each point on the graph of $y = - 2 \sqrt[3]{x + 3}$ :
Figure 4.11 $y = - 2 \sqrt[3]{x + 3}$
$→ add 1 to each y -value$
Figure 4.12 $y = - 2 \sqrt[3]{x + 3} + 1$
$(- 4, 2), (- 3, 0), (- 2, - 2)$
$(- 4, 3), (- 3, 1), (- 2, - 1)$
We get the domain and range of $f$ are $open parenthesis minus ∞, ∞ close parenthesis$ .
For $g (t) = \frac{\sqrt{1 - 2 t}}{4} = \frac{1}{4} \sqrt{- 2 t + 1}$ , we identify $n equals 2$ , $a = \frac{1}{4}$ , $b = - 2$ , $h = - 1$ and $k equals 0$ . Since we are asked to label three points on the graph, we track $open parenthesis 4, 2 close parenthesis$ along with $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 1, 1 close parenthesis$ .⁸
Step 1: add $minus 1$ to each of the $t$ -coordinates of each of the points on the graph of $y = \sqrt{t}$ :
Figure 4.13 $y = \sqrt{t}$
$→ add minus 1 to each t -coordinate$
Figure 4.14 $y = \sqrt{t + 1}$
$open parenthesis 0, 0 close parenthesis, open parenthesis 1, 1 close parenthesis, open parenthesis 4, 2 close parenthesis$
$open parenthesis minus 1, 0 close parenthesis, open parenthesis 0, 1 close parenthesis, open parenthesis 3, 2 close parenthesis$
Step 2: divide each of the $t$ -coordinates of each of the points on the graph of $y = \sqrt{t + 1}$ by $minus 2$ :
Figure 4.15 $y = \sqrt{t + 1}$
$\overset{divide each t -coordinate by - 2 :}{\overset{}{\to}}$
Figure 4.16 $y = \sqrt{- 2 t + 1}$
$open parenthesis minus 1, 0 close parenthesis, open parenthesis 0, 1 close parenthesis, open parenthesis 3, 2 close parenthesis$
$(\frac{1}{2}, 0), (0, 1), (- \frac{3}{2}, 2)$
Step 3: multiply each of the $y$ -coordinates of each of the points on the graph of $y = \sqrt{- 2 t + 1}$ by $\frac{1}{4}$ :
Figure 4.17 $y = \sqrt{- 2 t + 1}$
$\overset{multiply each y -coordinate by \frac{1}{4} :}{\overset{}{\to}}$
Figure 4.18 $y = \frac{1}{4} \sqrt{- 2 t + 1}$
$(\frac{1}{2}, 0), (0, 1), (- \frac{3}{2}, 2)$
$(\frac{1}{2}, 0), (0, \frac{1}{4}), (- \frac{3}{2}, \frac{1}{2})$
We get the domain is $(- \infty, \frac{1}{2}]$ and the range is $open bracket 0, ∞ close parenthesis$ .

Other Functions involving Radicals

Now that we have some practice with basic root functions, we turn our attention to more general functions involving radicals. In general, Calculus is the best tool with which to study these functions. Nevertheless, we will use what algebra we know in combination with a graphing utility to help us visualize these functions and preview concepts which are studied in greater depth in later courses. In the table below, we summarize some of the properties of radicals from elsewhere in this text (and Intermediate Algebra) we will be using in the coming examples.

Theorem

Some Useful Properties of Radicals: Suppose $\sqrt[n]{x}$ , $\sqrt[n]{a}$ , and $\sqrt[n]{b}$ are real numbers.⁹

Simplifying $n$ th powers and $n$ th roots: ¹⁰

${(\sqrt[n]{x})}^{n} = x$ .
if $n$ is odd, then $\sqrt[n]{x^{n}} = x$
if $n$ is even, then $\sqrt[n]{x^{n}} = | x |$ .

Root Functions Preserve Inequality: ¹¹ if $a \leq b$ , then $\sqrt[n]{a} \leq \sqrt[n]{b}$ .

Example 2

For the following functions:

Analytically:
- find the domain.
- find the axis intercepts.
- analyze the end behavior.
Graph the function with help from a graphing utility and determine:
- the range.
- the local extrema, if they exist.
- intervals of increase.
- intervals of decrease.
Construct a sign diagram for each function using the intercepts and graph.¹²

$f (x) = 3 x \sqrt[3]{2 - x}$
$g (t) = \sqrt[3]{\frac{8 t}{t + 1}}$
$h (x) = \frac{3 x}{\sqrt{x^{2} + 1}}$
$r (t) = t^{- 1} \sqrt{16 t^{4} - 1}$

When looking for the domain, we have two thing to watch out for: denominators (which we must make sure aren't $0$ ) and even indexed radicals (whose radicands we must ensure are nonnegative.) Looking at the expression for $f (x)$ , we have no denominators nor do we have an even indexed radical, so we are confident the domain is all real numbers, $open parenthesis minus ∞, ∞ close parenthesis$ .
To find the $x$ -intercepts, we find the zeros of $f$ by solving $f (x) = 3 x \sqrt[3]{2 - x} = 0$ . Using the zero product property, we get $3 ⁢ x equals 0$ or $\sqrt[3]{2 - x} = 0$ . The former gives $x equals 0$ and to solve the latter, we cube both sides and get $2 - x = 0$ or $x equals 2$ . Hence, the $x$ -intercepts are $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 2, 0 close parenthesis$ . Since $open parenthesis 0, 0 close parenthesis$ is also on the $y$ -axis and functions can can have at most one $y$ -intercept, we know $open parenthesis 0, 0 close parenthesis$ is the only $y$ -intercept.¹³ That being said, we can quickly verify $f (0) = 3 (0) \sqrt[3]{2 - 0} = 0$ .
To determine the end behavior, we first consider $f (x)$ as $x → ∞$ . Using `number sense,' ¹⁴ we have $f (x) = 3 x \sqrt[3]{2 - x} = 3 x \sqrt[3]{- x + 2} \approx (big (+)) \sqrt[3]{big (-)} = (big (+)) (big (-)) = big (-)$ , so $lim_{x \to \infty} f (x) = - \infty$ . As $x → minus ∞$ we get $f (x) = 3 x \sqrt[3]{- x + 2} \approx (big (-)) \sqrt[3]{big (+)} = (big (-)) (big (+)) = big (-)$ , so $lim_{x \to - \infty} f (x) = - \infty$ as well.
We graph $f$ below on the left. From the graph, the range appears to be $open parenthesis minus ∞, 3.572 close bracket$ with a local maximum (which also happens to be the maximum) at $open parenthesis 1.5, 3.572 close parenthesis$ . We also see $f$ appears to be increasing on $open parenthesis minus ∞, 1.5 close parenthesis$ and decreasing on $open parenthesis 1.5, ∞ close parenthesis$ . It is also worth noting that there appears to be `unusual steepness' near the $x$ -intercept $open parenthesis 2, 0 close parenthesis$ . We invite the reader to zoom in on the graph near $open parenthesis 2, 0 close parenthesis$ to see that the function is `locally vertical.'¹⁵
To create a sign diagram for $f (x)$ , we note that the function has zeros $x equals 0$ and $x equals 2$ . For $x less than 0$ , $f ⁢ open parenthesis x close parenthesis less than 0$ or $(-)$ , for $0 less than x less than 2$ , $f ⁢ open parenthesis x close parenthesis greater than 0$ or $(+)$ , and for $x greater than 2$ , $f ⁢ open parenthesis x close parenthesis less than 0$ or $(-)$ . The sign diagram for $f (x)$ is below on the right.
Figure 4.19
Figure 4.20
The graph of $y = f (x)$
Sign Diagram for $f (x)$
The index of the radical in the expression for $g (t)$ is odd, so our only concern is the denominator. Setting $t + 1 = 0$ gives $t = - 1$ , which we exclude, so our domain is $open brace t ∈ ℝ vertical bar t not equal to minus 1 close brace$ or using interval notation, $(- \infty, - 1) \cup (- 1, \infty)$ .
If we take the time to analyze the behavior of $g$ near $t = - 1$ , we find that as $t \to - 1^{-}$ , $g (t) = \sqrt[3]{\frac{8 t}{t + 1}} \approx \sqrt[3]{\frac{- 8}{small (-)}} \approx \sqrt[3]{big (+)} = big (+)$ . That is, $lim_{t \to - 1^{-}} g (t) = \infty$ . Likewise, as $t \to - 1^{+}$ , $g (t) \approx \sqrt[3]{\frac{- 8}{small (+)}} \approx \sqrt[3]{big (-)} = big (-)$ . This suggests $lim_{t \to - 1^{+}} g (t) = - \infty$ . This behavior points to a vertical asymptote, $t = - 1$ .
To find the $t$ -intercepts of the graph of $g$ , we find the zeros of $g$ by setting $g (t) = \sqrt[3]{\frac{8 t}{t + 1}} = 0$ . Cubing both sides and clearing denominators gives $8 ⁢ t equals 0$ or $t equals 0$ . Hence our $t$ -, and in this case, $y$ - intercept is $open parenthesis 0, 0 close parenthesis$ .
To determine the end behavior, we note that as $t → minus ∞$ or $t → ∞$ , $\frac{8 t}{t + 1} \approx \frac{8 t}{t} = 8$ . Since $g (t) = \sqrt[3]{\frac{8 t}{t + 1}}$ it stands to reason that $lim_{t \to - \infty} g (t) = \sqrt[3]{8} = 2$ and, likewise, $lim t → ∞ g ⁢ open parenthesis t close parenthesis equals 2$ This suggests the graph of $y = g (t)$ has a horizontal asymptote at $y equals 2$ .
We graph $y = g (t)$ below on the left. The graph confirms our suspicions about the asymptotes $t = - 1$ and $y equals 2$ . Moreover, the range appears to be $open parenthesis minus ∞, 2 close parenthesis union open parenthesis 2, ∞ close parenthesis$ .
We could check if the graph ever crosses its horizontal asymptote by attempting to solve $g (t) = \sqrt[3]{\frac{8 t}{t + 1}} = 2$ . Cubing both sides and clearing denominators gives $8 t = 8 (t + 1)$ which gives $0 equals 8$ , a contradiction. This proves $2$ is not in the range, as we had suspected.
Scanning the graph, there appears to be no local extrema, and, moreover, the graph suggests $g$ is increasing on $(- \infty, - 1)$ and again on $open parenthesis minus 1, ∞ close parenthesis$ . As with the previous example, the graph appears locally vertical near its intercept $open parenthesis 0, 0 close parenthesis$ .
To create a sign diagram for $g (t)$ , we note that the function is undefined when $t = - 1$ (so we place a `‽' above it) and has a zero $t equals 0$ . When $t less than minus 1$ , $g ⁢ open parenthesis t close parenthesis greater than 0$ or $(+)$ , for $minus 1 less than t less than 0$ , $g ⁢ open parenthesis t close parenthesis less than 0$ or $(-)$ , and for $t greater than 0$ , $g ⁢ open parenthesis t close parenthesis greater than 0$ or $(+)$ . Below on the right is a sign diagram for $g (t)$ .
Figure 4.21
Figure 4.22
The graph of $y = g (t)$
Sign Diagram for $g (t)$
The expression for $h (x) = \frac{3 x}{\sqrt{x^{2} + 1}}$ has both a denominator and an even-indexed radical, so we have to be extra cautious here. Fortunately for us, the quantity $x squared plus 1 greater than 0$ for all real numbers $x$ . Not only does this mean $\sqrt{x^{2} + 1}$ is always defined, it also tells us $\sqrt{x^{2} + 1} > 0$ for all $x$ , too. This means the domain of $h$ is all real numbers, $open parenthesis minus ∞, ∞ close parenthesis$ .
Solving for the zeros of $h$ gives only $x equals 0$ , and we find, once again, $open parenthesis 0, 0 close parenthesis$ is both our lone $x$ - and $y$ -intercept.
Moving on to end behavior, as $x → minus ∞$ or $x → ∞$ , the term $x^{2}$ is the dominant term in the radicand in the denominator. As such, $h (x) = \frac{3 x}{\sqrt{x^{2} + 1}} \approx \frac{3 x}{\sqrt{x^{2}}} = \frac{3 x}{| x |}$ . As $x → minus ∞$ , $| x | = - x$ (since $x less than 0$ ) and hence, $h (x) \approx \frac{3 x}{- x} = - 3$ , so $lim_{x \to - \infty} h (x) = - 3$ . As $x → ∞$ , $| x | = x$ (since $x greater than 0$ ), so $h (x) \approx \frac{3 x}{x} = 3$ , so $lim x → ∞ h ⁢ open parenthesis x close parenthesis equals 3$ .
This analysis suggests the graph of $y = h (x)$ has not one, but two horizontal asymptotes.¹⁶ The graph of $h$ below on the left bears this out.
From the graph, we see the range of $h$ appears to be $open parenthesis minus 3, 3 close parenthesis$ . Attempting to solve $h (x) = \frac{3 x}{\sqrt{x^{2} + 1}} = - 3$ or $h (x) = \frac{3 x}{\sqrt{x^{2} + 1}} = 3$ gives, in either case, $9 x^{2} = 9 (x^{2} + 1)$ which reduces to $0 equals 9$ , a contradiction. Hence, the graph of $y = h (x)$ never reaches its horizontal asymptotes. Moreover, $h$ appears to be always increasing, with no local extrema or `unusual' steepness. One last remark: it appears as if the graph of $h$ is symmetric about the origin. We check $h (- x) = \frac{3 (- x)}{\sqrt{{(- x)}^{2} + 1}} = - \frac{3 x}{\sqrt{x^{2} + 1}} = - h (x)$ which verifies $h$ is odd.
Since the domain of $h$ is all real number and the only zero of $h$ is $x equals 0$ , the sign diagram for $h (x)$ is fairly straight forward. For $x less than 0$ , $h ⁢ open parenthesis x close parenthesis less than 0$ or $(-)$ and for $x greater than 0$ , $h ⁢ open parenthesis x close parenthesis greater than 0$ or $(+)$ . The sign diagram for $h (x)$ is below on the right.
Figure 4.23
Figure 4.24
The graph of $y = h (x)$
Sign Diagram for $h (x)$
The first thing to note about the expression $r (t) = t^{- 1} \sqrt{16 t^{4} - 1}$ is that $t^{- 1} = \frac{1}{t}$ . Hence, we must exclude $t equals 0$ from the domain straight away. Next, we have an even-indexed radical expression: $\sqrt{16 t^{4} - 1}$ . In order for this to return a real number, we require $16 t^{4} - 1 \geq 0$ . Instead of using a sign diagram to solve this,¹⁷ we opt instead to carefully use properties of radicals. Isolating $t to the power 4$ , we have $t^{4} \geq \frac{1}{16}$ . Since the root functions are increasing, we can apply the fourth root to both sides and preserve the inequality: $\sqrt[4]{t^{4}} \geq \sqrt[4]{\frac{1}{16}}$ which gives¹⁸ $| t | \geq \frac{1}{2}$ . Note that since $t equals 0$ does not satisfy this inequality, restricting $t$ in this manner takes care of both domain issues, so the domain is $(- \infty, - \frac{1}{2}] \cup [\frac{1}{2}, \infty)$ .
Next, we look for zeros. Setting $r (t) = t^{- 1} \sqrt{16 t^{4} - 1} = \frac{\sqrt{16 t^{4} - 1}}{t} = 0$ gives $\sqrt{16 t^{4} - 1} = 0$ . After squaring both sides, we get $16 t^{4} - 1 = 0$ or $t^{4} = \frac{1}{16}$ . Extracting fourth roots, we get $t = \pm \frac{1}{2}$ . Both of these are (barely!) in the domain of $r$ , so our $t$ intercepts are $(- \frac{1}{2}, 0)$ and $(\frac{1}{2}, 0)$ . Note, the graph of $r$ has no $y$ -intercept, since $r ⁢ open parenthesis 0 close parenthesis$ is undefined ( $t equals 0$ is not in the domain of $r$ ).
Concerning end behavior, we note the term $16 ⁢ t to the power 4$ dominates the radicand $\sqrt{16 t^{4} - 1}$ as $t → minus ∞$ or $t → ∞$ , hence, $r (t) = \frac{\sqrt{16 t^{4} - 1}}{t} \approx \frac{\sqrt{16 t^{4}}}{t} = \frac{4 t^{2}}{t} = 4 t$ . This suggests the graph of $y = r (t)$ has a slant asymptote with slope $4$ .¹⁹ At this point, we can at least write $lim_{t \to - \infty} r (t) = - \infty$ . and $lim t → ∞ r ⁢ open parenthesis t close parenthesis equals ∞$ .
We graph $y = r (t)$ below on the left. We see the range appears to be all real numbers, $open parenthesis minus ∞, ∞ close parenthesis$ . It appears as if $r$ is increasing on $(- \infty, - \frac{1}{2}]$ and again on $[\frac{1}{2}, \infty)$ . The graph does appear to be asymptotic to $y equals 4 ⁢ t$ , and it also appears to be symmetric about the origin. Sure enough, we find $r (- t) = \frac{\sqrt{16 {(- t)}^{4} - 1}}{- t} = - \frac{\sqrt{16 t^{4} - 1}}{t} = - r (t)$ , proving $r$ is an odd function.
To construct the sign diagram for $r (t)$ we note $r$ has two zeros, $t = \pm \frac{1}{2}$ . For $t < \frac{1}{2}$ , $r ⁢ open parenthesis t close parenthesis less than 0$ or $(-)$ and when $t > \frac{1}{2}$ , $r ⁢ open parenthesis t close parenthesis greater than 0$ or $(+)$ . When $- \frac{1}{2} < t < \frac{1}{2}$ , $r$ is undefined so we have removed that segment from the diagram, as seen below on the right.
Figure 4.25
Figure 4.26
The graph of $y = r (t)$
Sign Diagram for $r (t)$ .

We end this section with a classic application of root functions.

Example 3

Carl wishes to get high speed internet service installed in his remote Sasquatch observation post located $30$ miles from Route $117$ . The nearest junction box is located $50$ miles down the road from the post, as indicated in the diagram below. Suppose it costs $$ 15$ per mile to run cable along the road and $$ 20$ per mile to run cable off of the road.

Find an expression $C (x)$ which computes the cost of connecting the Junction Box to the Outpost as a function of $x$ , the number of miles the cable is run along Route $117$ before heading off road directly towards the Outpost. Determine a reasonable applied domain for the problem.
Use your calculator to graph $y = C (x)$ on its domain. What is the minimum cost? How far along Route $117$ should the cable be run before turning off of the road?

Figure 4.27

The cost is broken into two parts: the cost to run cable along Route $117$ at $$ 15$ per mile, and the cost to run it off road at $$ 20$ per mile. Since $x$ represents the miles of cable run along Route $117$ , the cost for that portion is $15 ⁢ x$ . From the diagram, we see that the number of miles the cable is run off road is $z$ , so the cost of that portion is $20 ⁢ z$ . Hence, the total cost is $15 ⁢ x plus 20 ⁢ z$ .
Our next goal is to determine $z$ in terms of $x$ . The diagram suggests we can use the Pythagorean Theorem to get $y^{2} + 30^{2} = z^{2}$ . But we also see $x + y = 50$ so that $y = 50 - x$ . Substituting $open parenthesis 50 minus x close parenthesis$ in for $y$ we obtain $z^{2} = {(50 - x)}^{2} + 900$ . Solving for $z$ , we obtain $z = \pm \sqrt{{(50 - x)}^{2} + 900}$ . Since $z$ represents a distance, we choose $z = \sqrt{{(50 - x)}^{2} + 900}$ .
Hence, the cost as a function of $x$ is given by $C (x) = 15 x + 20 \sqrt{{(50 - x)}^{2} + 900}$ . From the context of the problem, we have $0 less than or equal to x less than or equal to 50$ .
We graph $y = C (x)$ below and find our (local) minimum to be at the point $open parenthesis 15.98, 1146.86 close parenthesis$ . Here the $x$ -coordinate tells us that in order to minimize cost, we should run $15.98$ miles of cable along Route 117 and then turn off of the road and head towards the outpost. The $y$ -coordinate tells us that the minimum cost, in dollars, to do so is $$ 1146.86$ . The ability to stream live SasquatchCasts? Priceless.
Figure 4.28

Exercises

In Exercises -, given the pair of functions $f$ and $F$ , sketch the graph of $y = F (x)$ by starting with the graph of $y = f (x)$ and using Theorem 4.1. Track at least two points and state the domain and range using interval notation.

$f (x) = \sqrt{x}$ , $F (x) = \sqrt{x + 3} - 2$
$f (x) = \sqrt{x}$ , $F (x) = \sqrt{4 - x} - 1$
$f (x) = \sqrt[3]{x}$ , $F (x) = \sqrt[3]{x - 1} - 2$
$f (x) = \sqrt[3]{x}$ , $F (x) = - \sqrt[3]{8 x + 8} + 4$
$f (x) = \sqrt[4]{x}$ , $F (x) = \sqrt[4]{x - 1} - 2$
$f (x) = \sqrt[4]{x}$ , $F (x) = - 3 \sqrt[4]{x - 7} + 1$
$f (x) = \sqrt[5]{x}$ , $F (x) = \sqrt[5]{x + 2} + 3$
$f (x) = \sqrt[8]{x}$ , $F (x) = \sqrt[8]{- x} - 2$

In Exercises -, find a formula for each function below in the form $F (x) = a \sqrt{b x - h} + k$ .

NOTE: There may be more than one solution!

$y = F (x)$
Figure 4.29 $x$ , $y$ -intercept $open parenthesis 0, 0 close parenthesis$
$y = F (x)$
Figure 4.30 $x$ -intercept $open parenthesis 1, 0 close parenthesis$ , $y$ -intercept $open parenthesis 0, 2 close parenthesis$

In Exercises -, find a formula for each function below in the form $F (x) = a \sqrt[3]{b x - h} + k$ .

NOTE: There may be more than one solution!

$y = F (x)$
Figure 4.31 $x$ -intercept $(- \frac{1}{2}, 0)$ , $y$ -intercept $open parenthesis 0, minus 1 close parenthesis$
$y = F (x)$
Figure 4.32 $x$ -intercept $open parenthesis 2, 0 close parenthesis$ , $y$ -intercept $open parenthesis 0, minus 4 close parenthesis$
Use the fact that the $n$ th root functions are increasing to solve the following polynomial inequalities:
1. $x cubed less than or equal to 64$
2. $2 - t^{5} < 34$
3. $\frac{{(2 z + 1)}^{3}}{4} \geq 2$
For the following inequalities, remember $\sqrt[n]{x^{n}} = | x |$ if $n$ is even:
1. $x to the power 4 less than or equal to 16$
2. $6 - t^{6} < - 58$
3. $\frac{{(2 z + 1)}^{4}}{3} \geq 27$

For each function in Exercises - below

Analytically:
- find the domain.
- find the axis intercepts.
- analyze the end behavior.
Graph the function with help from a graphing utility and determine:
- the range.
- the local extrema, if they exist.
- intervals of increase/decrease.
- any `unusual steepness' or `local' verticality.
- vertical asymptotes.
- horizontal / slant asymptotes.
Construct a sign diagram for each function using the intercepts and graph.
Comment on any observed symmetry.
$f (x) = \sqrt{1 - x^{2}}$
$f (x) = \sqrt{x^{2} - 1}$
$g (t) = t \sqrt{1 - t^{2}}$
$g (t) = t \sqrt{t^{2} - 1}$
$f (x) = \sqrt[4]{\frac{16 x}{x^{2} - 9}}$
$f (x) = \frac{5 x}{\sqrt[3]{x^{3} + 8}}$
$g (t) = \sqrt{t (t + 5) (t - 4)}$
$g (t) = \sqrt[3]{t^{3} + 3 t^{2} - 6 t - 8}$
Rework Example 4.1.3 so that the outpost is 10 miles from Route 117 and the nearest junction box is 30 miles down the road for the post.
The volume $V$ of a right cylindrical cone depends on the radius of its base $r$ and its height $h$ and is given by the formula $V = \frac{1}{3} π r^{2} h$ . The surface area $S$ of a right cylindrical cone also depends on $r$ and $h$ according to the formula $S = π r \sqrt{r^{2} + h^{2}}$ . In the following problems, suppose a cone is to have a volume of 100 cubic centimeters.
1. Use the formula for volume to find the height as a function of $r$ , $h (r)$ .
2. Use the formula for surface area along with your answer to to find the surface area as a function of $r$ , $S (r)$ .
3. Use your calculator to find the values of $r$ and $h$ which minimize the surface area. What is the minimum surface area? Round your answers to two decimal places.
The period of a pendulum in seconds is given by
$T = 2 π \sqrt{\frac{L}{g}}$
(for small displacements) where $L$ is the length of the pendulum in meters and $g equals 9.8$ meters per second per second is the acceleration due to gravity. My Seth-Thomas antique schoolhouse clock needs $T = \frac{1}{2}$ second and I can adjust the length of the pendulum via a small dial on the bottom of the bob. At what length should I set the pendulum?
According to Einstein's Theory of Special Relativity, the observed mass of an object is a function of how fast the object is traveling. Specifically, if $m_{r}$ is the mass of the object at rest, $v$ is the speed of the object and $c$ is the speed of light, then the observed mass of the object $m (v)$ is given by:
$m (v) = \frac{m_{r}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$
1. Find the applied domain of the function.
2. Compute $m ⁢ open parenthesis .1 ⁢ c close parenthesis, m ⁢ open parenthesis .5 ⁢ c close parenthesis, m ⁢ open parenthesis .9 ⁢ c close parenthesis$ and $m ⁢ open parenthesis .999 ⁢ c close parenthesis$ .
3. Find $lim_{v \to c^{-}} m (v)$ .
4. How slowly must the object be traveling so that the observed mass is no greater than 100 times its mass at rest?
Find the inverse of $k (x) = \frac{2 x}{\sqrt{x^{2} - 1}}$ .

Answers

$F (x) = \sqrt{x + 3} - 2$
Figure 4.33 Domain: $open bracket minus 3, ∞ close parenthesis$ , Range: $open bracket minus 2, ∞ close parenthesis$
$F (x) = \sqrt{4 - x} - 1 = \sqrt{- x + 4} - 1$
Figure 4.34 Domain: $open parenthesis minus ∞, 4 close bracket$ , Range: $open bracket minus 1, ∞ close parenthesis$
$F (x) = \sqrt[3]{x - 1} - 2$
Figure 4.35 Domain: $open parenthesis minus ∞, ∞ close parenthesis$ , Range: $open parenthesis minus ∞, ∞ close parenthesis$
$F (x) = - \sqrt[3]{8 x + 8} + 4$
Figure 4.36 Domain: $open parenthesis minus ∞, ∞ close parenthesis$ , Range: $open parenthesis minus ∞, ∞ close parenthesis$
$F (x) = \sqrt[4]{x - 1} - 2$
Figure 4.37 Domain: $open bracket 1, ∞ close parenthesis$ , Range: $open bracket minus 2, ∞ close parenthesis$
$F (x) = - 3 \sqrt[4]{x - 7} + 1$
Figure 4.38 Domain: $open bracket 7, ∞ close parenthesis$ , Range: $open parenthesis minus ∞, 1 close bracket$
$F (x) = \sqrt[5]{x + 2} + 3$
Figure 4.39 Domain: $open parenthesis minus ∞, ∞ close parenthesis$ , Range: $open parenthesis minus ∞, ∞ close parenthesis$
$F (x) = \sqrt[8]{- x} - 2$
Figure 4.40 Domain: $open parenthesis minus ∞, 0 close bracket$ , Range: $open bracket minus 2, ∞ close parenthesis$
One solution is: $F (x) = - \sqrt{x + 4} + 2$
One solution is: $F (x) = 2 \sqrt{- x + 1}$
One solution is: $F (x) = - \sqrt[3]{2 x + 1}$
One solution is: $F (x) = 2 \sqrt[3]{x - 1} - 2$
1. $open parenthesis minus ∞, 4 close bracket$
2. $open parenthesis minus 2, ∞ close parenthesis$
3. $[\frac{1}{2}, \infty)$
1. $open bracket minus 2, 2 close bracket$
2. $(- \infty, - 2) \cup (2, \infty)$
3. $(- \infty, - 2] \cup [1, \infty)$
$f (x) = \sqrt{1 - x^{2}}$ Domain: $open bracket minus 1, 1 close bracket$ Intercepts: $open parenthesis minus 1, 0 close parenthesis$ , $open parenthesis 1, 0 close parenthesis$ Range: $open bracket 0, 1 close bracket$ Local maximum: $open parenthesis 0, 1 close parenthesis$ Increasing: $open bracket minus 1, 0 close bracket$ , Decreasing: $open bracket 0, 1 close bracket$ Unusual steepness²⁰ at $x = - 1$ and $x equals 1$ Sign Diagram:
Figure 4.41
Graph:
Figure 4.42
Note: $f$ is even.
$f (x) = \sqrt{x^{2} - 1}$ Domain: $(- \infty, - 1] \cup [1, \infty)$ Intercepts: $open parenthesis minus 1, 0 close parenthesis$ , $open parenthesis 1, 0 close parenthesis$ $lim_{x \to - \infty} f (x) = \infty$ ²¹ $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals ∞$ Range: $open bracket 0, ∞ close parenthesis$ Increasing: $open bracket 1, ∞ close parenthesis$ , Decreasing: $(- \infty, - 1]$ Unusual steepness²²at $x = - 1$ and $x equals 1$ Sign Diagram:
Figure 4.43
Graph:
Figure 4.44
Note: $f$ is even.
$g (t) = t \sqrt{1 - t^{2}}$ Domain: $open bracket minus 1, 1 close bracket$ Intercepts: $open parenthesis minus 1, 0 close parenthesis$ , $open parenthesis 0, 0 close parenthesis$ , $open parenthesis 1, 0 close parenthesis$ Range: $\approx [- 0.5, 0.5]$ Local minimum $\approx (- 0.707, - 0.5)$ Local maximum: $approximately equals open parenthesis 0.707, 0.5 close parenthesis$ Increasing: $\approx [- 0.707, 0.707]$ Decreasing: $\approx [- 1, - 0.707]$ , $open bracket 0.707, 1 close bracket$ Unusual steepness at $t = - 1$ and $t equals 1$ Sign Diagram:
Figure 4.45
Graph:
Figure 4.46
Note: $g$ is odd.
$g (t) = t \sqrt{t^{2} - 1}$ Domain: $(- \infty, - 1] \cup [1, \infty)$ Intercepts: $open parenthesis minus 1, 0 close parenthesis$ , $open parenthesis 1, 0 close parenthesis$ $lim_{t \to - \infty} g (t) = - \infty$ $lim t → ∞ g ⁢ open parenthesis t close parenthesis equals ∞$ Range: $open parenthesis minus ∞, ∞ close parenthesis$ Increasing: $(- \infty, - 1]$ , $open bracket 1, ∞ close parenthesis$ Unusual steepness at $t = - 1$ and $t equals 1$ Sign Diagram:
Figure 4.47
Graph:
Figure 4.48
Note: $g$ is odd.
$f (x) = \sqrt[4]{\frac{16 x}{x^{2} - 9}}$ Domain: $open parenthesis minus 3, 0 close bracket union open parenthesis 3, ∞ close parenthesis$ Intercept: $open parenthesis 0, 0 close parenthesis$ Range: $open bracket 0, ∞ close parenthesis$ Decreasing: $open parenthesis minus 3, 0 close bracket$ , $open parenthesis 3, ∞ close parenthesis$ Unusual steepness at $x equals 0$ Vertical asymptotes: $x = - 3$ and $x equals 3$ Horizontal asymptote: $y equals 0$ Sign Diagram:
Figure 4.49
Graph:
Figure 4.50
$f (x) = \frac{5 x}{\sqrt[3]{x^{3} + 8}}$ Domain: $(- \infty, - 2) \cup (- 2, \infty)$ Intercept: $open parenthesis 0, 0 close parenthesis$ Range: $open parenthesis minus ∞, 5 close parenthesis union open parenthesis 5, ∞ close parenthesis$ Increasing: $(- \infty, - 2)$ , $open parenthesis minus 2, ∞ close parenthesis$ Vertical asymptote $x = - 2$ Horizontal asymptote $y equals 5$ Sign Diagram:
Figure 4.51
Graph:
Figure 4.52
$g (t) = \sqrt{t (t + 5) (t - 4)}$ Domain: $open bracket minus 5, 0 close bracket union open bracket 4, ∞ close parenthesis$ Intercepts $open parenthesis minus 5, 0 close parenthesis$ , $open parenthesis 0, 0 close parenthesis$ , $open parenthesis 4, 0 close parenthesis$ $lim t → ∞ g ⁢ open parenthesis t close parenthesis equals ∞$ Range: $open bracket 0, ∞ close parenthesis$ Local maximum $\approx (- 2.937, 6.483)$ Increasing: $\approx [- 5, - 2.937]$ , $open bracket 4, ∞ close parenthesis$ Decreasing: $\approx [- 2.937, 0]$ Unusual steepness at $t = - 5, t = 0$ and $t equals 4$ Sign Diagram:
Figure 4.53
Graph:
Figure 4.54
$g (t) = \sqrt[3]{t^{3} + 3 t^{2} - 6 t - 8}$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$ Intercepts: $open parenthesis minus 4, 0 close parenthesis$ , $open parenthesis minus 1, 0 close parenthesis$ , $open parenthesis 0, minus 2 close parenthesis$ , $open parenthesis 2, 0 close parenthesis$ ²³ $lim_{t \to - \infty} g (t) = - \infty$ $lim t → ∞ g ⁢ open parenthesis t close parenthesis equals ∞$ Range: $open parenthesis minus ∞, ∞ close parenthesis$ Local maximum: $\approx (- 2.732, 2.182)$ Local minimum: $\approx (0.732, - 2.182)$ Increasing: $\approx (- \infty, - 2.732]$ , $open bracket 0.732, ∞ close parenthesis$ Decreasing: $\approx [- 2.732, 0.732]$ Unusual steepness at $t = - 4, t = - 1$ and $t equals 2$
Sign Diagram:
Figure 4.55
Graph:
Figure 4.56
$C (x) = 15 x + 20 \sqrt{100 + {(30 - x)}^{2}}$ , $0 less than or equal to x less than or equal to 30$ . The calculator gives the absolute minimum at approximately $open parenthesis 18.66, 582.29 close parenthesis$ . This means to minimize the cost, approximately 18.66 miles of cable should be run along Route 117 before turning off the road and heading towards the outpost. The minimum cost to run the cable is approximately $$ 582.29$ .
1. $h (r) = \frac{300}{π r^{2}}$ , $r greater than 0$ .
2. $S (r) = π r \sqrt{r^{2} + {(\frac{300}{π r^{2}})}^{2}} = \frac{\sqrt{π^{2} r^{6} + 90000}}{r}$ , $r greater than 0$
3. The calculator gives the absolute minimum at the point $approximately equals open parenthesis 4.07, 90.23 close parenthesis$ . This means the radius should be (approximately) 4.07 centimeters and the height should be 5.76 centimeters to give a minimum surface area of 90.23 square centimeters.
$9.8 {(\frac{1}{4 π})}^{2} \approx 0.062$ meters or $6.2$ centimeters
1. $open bracket 0, c close parenthesis$
2. $m (.1 c) = \frac{m_{r}}{\sqrt{.99}} \approx 1.005 m_{r}$ , $m (.5 c) = \frac{m_{r}}{\sqrt{.75}} \approx 1.155 m_{r}$ , $m (.9 c) = \frac{m_{r}}{\sqrt{.19}} \approx 2.294 m_{r}$ , $m (.999 c) = \frac{m_{r}}{\sqrt{.0.001999}} \approx 22.366 m_{r}$ .
3. $lim_{v \to c^{-}} m (x) \to \infty$ ; as the object's velocity approaches the speed of light, mass becomes infinite.
4. If the object is traveling no faster than approximately $0.99995$ times the speed of light, then its observed mass will be no greater than $100 ⁢ m sub r$ .
$k^{- 1} (x) = \frac{x}{\sqrt{x^{2} - 4}}$

These were introduced, as you may recall, as piecewise-defined linear functions. ↩
This is a really bad math pun. ↩
If we broaden our concept of functions to allow for complex valued coefficients, the Complex Factorization Theorem, Theorem 2.16, tells us every function we have studied thus far is a combination of linear functions. ↩
Although we discussed imaginary numbers in Section 2.4, we restrict our attention to real numbers in this section. See the epilogue on page the related section for more details. ↩
See Exercise . ↩
Since, otherwise, $- 1 = i^{2} = i \cdot i = \sqrt{- 1} \sqrt{- 1} = \sqrt{(- 1) (- 1)} = \sqrt{1} = 1$ , a contradition. ↩
again this is because every real number can be represented as both $x - h$ for some value $x$ and as $c + h$ for some value $c$ . ↩
As $\sqrt{4} = 2$ , we know $open parenthesis 4, 2 close parenthesis$ is on the graph of $y = \sqrt{t}$ . ↩
i.e., if $n$ is odd, $x$ , $a$ , and $b$ can be any real numbers; if, on the other hand $n$ is even, $x greater than or equal to 0$ , $a greater than or equal to 0$ , and $b greater than or equal to 0$ . ↩
a.k.a., `Inverse Properties.' See Section 5.6. ↩
i.e., root functions are increasing. ↩
We'll revisit sign diagrams for these functions in Section 4.3 where we will use them to solve inequalities (surprised?) ↩
Why is this, again? ↩
remember this means we use the adjective `big' here to mean large in absolute value ↩
Of course, the Vertical Line Test prohibits the graph from actually being a vertical line. This behavior is more precisely defined and more closely studied in Calculus. ↩
We warned you this was coming …see the discussion following Theorem 3.3 in Section 3.1. ↩
See Section 2.3 ↩
Recall: $\sqrt[n]{x^{n}} = | x |$ , not $x$ , if $n$ is even. ↩
Note: this analysis suggests the slant asymptote is $y = 4 t + b$ , but from this analysis, we cannot determine the value of $b$ . As with slant asymptotes in Section 3.1, we'd need to perform a more detailed analysis which we omit in this case owing to the complexity of the function. (You'll have the tools in Calculus, however!) ↩
You may need to zoom in to see this. ↩
Using Calculus, one can show $y = - x$ and $y = - x$ are slant asymptotes to the graph. ↩
You may need to zoom in to see this. ↩
Using Calculus it can be shown that $y = t + 1$ is a slant asymptote of this graph. ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.