Graphing Rational Functions

Notice that there is no $x\text{-}$value that can make $x^2+1$ equal zero. However, the other factor in the denominator makes the function undefined when $x=\text{-}1.$ Therefore, this value is excluded from the domain. Recall that the point of discontinuity occurs at $x=a$ when $a$ is not in the domain of a function. Consequently, there is a point of discontinuity at $x=\text{-}1.$ In the context of the problem, this is the $x\text{-}$coordinate of the point where the road is being repaired.

Notice that it is also a removable discontinuity because $x+1$ is the common factor in the numerator and denominator.

The function is undefined when $x=\text{-}1.$ This means that the domain of the function is all real numbers except $x=\text{-}1.$ After simplification, the function will be a linear function except at $x=\text{-}1.$ Note that $\frac{2x-3}{10}=\frac{2x}{10}-\frac{3}{10}.$ Therefore, the graph of the given function is the graph of the function $y=\frac{2x}{10}-\frac{3}{10}$ with a hole at $x=\text{-}1.$

This graph also has a removable discontinuity at $x=\text{-}1$ because $x+1$ is the common factor in the numerator and denominator of the rational function.

To identify the asymptotes, the numerator and denominator of the function $A(t)$ need to be factored. Notice that the terms in the numerator have a common factor of $6t.$ To factor the polynomial in the denominator, use the Factor Theorem.

Factor Theorem

The expression $x-a$ is a factor of a polynomial if and only if the value $a$ is a zero of the related polynomial function.

Now, the aim is to find all real zeros of the polynomial. Before doing so, identify the degree of the polynomial. It is a polynomial with a degree of $3.$ Subsequently, by the Fundamental Theorem of Algebra, it is known know that $p(t)$ has exactly three roots. Additionally, by the Rational Root Theorem, integer roots must be factors of the constant term. Since the constant term of $p(t)$ is $\text{-}8,$ the possible integer roots are $\pm 1,$ $\pm 2,$ $\pm 4,$ and $\pm 8.$

$t$	$t^3-t^2-10t-8$	$p(t)=t^3-t^2-10t-8$
$1$	$1^3-(1{)}^2-10(1)-8$	$\text{-}18$
$\text{-}1$	${\text{-}1}^3-(\text{-}1{)}^2-10(\text{-}1)-8$	$0$
$2$	$2^3-(2{)}^2-10(2)-8$	$\text{-}24$
$\text{-}2$	${\text{-}2}^3-(\text{-}2{)}^2-10(\text{-}2)-8$	$0$
$4$	$4^3-6(4{)}^2-7(4)+60$	$0$

Three roots for $p(t)$ were found, so there is no need for further investigation. In conclusion, $\text{-}1,$ $\text{-}2,$ and $4$ are all the real roots of $t^3-t^2-10t-8.$ Next, apply the Factor Theorem to write its factored form. Now that the factored form of the denominator has been found, the given function can be simplified. Since the numerator and the denominator share the factor $t-4,$ there is a hole at $t=4.$ Also, the denominator becomes zero when $t=\text{-}1$ and $t=\text{-}2.$ These values make the function undefined. Therefore, the function has two vertical asymptotes. To find the horizontal asymptote, compare the degrees of the polynomials in the numerator and denominator.

$y=\frac{a{x}^m}{b{x}^n}$	Asymptote
$m<n$	$y=0$
$m>n$	none
$m=n$	$y=\frac{a}{b}$

Look at the degrees of the numerator and denominator of the function. The degree of the denominator is greater than the degree of the numerator, $m<n.$ Therefore, the horizontal asymptote is $y=0.$

First, the rational expressions on the right-hand side will be subtracted. The least common denominator of the expressions is $2(t+1{)}^2.$ Therefore, the first expression needs to be multiplied by $\frac{2(t+1)}{2(t+1)}$ so that both have the same denominator. The numerator and the denominator do not have any common factors. Therefore, the function does not have a point of discontinuity, or hole, in its graph. However, it has a vertical asymptote at $t=\text{-}1$ because the function is undefined at that value. Now, to determine if the given rational function has a horizontal or oblique asymptote, the degrees of the polynomials in the numerator and the denominator will be compared. The degree of the numerator is greater than the degree of the denominator, so the graph has no horizontal asymptotes. However, since the degree of the numerator is $1$ more than that of the denominator, the graph has an oblique asymptote. Its equation is found by dividing the numerator by the denominator. Use long division to do this. The given function can be rewritten as follows. Therefore, the equation of the oblique asymptote is $O(t)=\text{-}\frac{t}{2}+2.$

The surface area of a rectangular box is the total area of all the surfaces of the shape. Since the dimensions are given in terms of $x,$ the surface area is a function of $x.$ Next, the volume of the pencil box will be calculated. Recall that the volume of a rectangular prism is the product of its dimensions. Finally, the function $E(x),$ the ratio of $S(x)$ to $V(x),$ can be written.

Asymptotes

Consider the function written in the previous part. First, factor the denominator of this function to identify its domain. The rational function is undefined where $x-2=0$ and $x+3=0$ because division by zero is not defined. This means that $x=2$ and $x=\text{-}3$ are not included in the domain. Asymptotes can be vertical, horizontal, or slanted lines. Once again, consider the function. It seems that the denominator and the numerator do not have a common factor. To find the factors of the numerator, its zeros can be found using the Quadratic Formula. As shown, they do not have common factors. Therefore, there are no holes in the graph. Recall that if the real number $a$ is not included in the domain, there is a vertical asymptote at $x=a.$ In this case, there are two vertical asymptotes, one at $x=\text{-}3$ and the other at $x=2.$ To find whether the graph has a horizontal or oblique asymptote, review how its type is determined using the degrees of the polynomials in the numerator and the denominator.

$y=\frac{a{x}^m+\dots }{b{x}^n+\dots }$	Asymptote
$m<n$	$y=0$
$m>n$	none
$m=n$	$y=\frac{a}{b}$
$m-n=1$	$y=$ the quotient of the polynomials with no remainder

Look at the degrees of the numerator and denominator of the function. The degree of the denominator is equal to the degree of the numerator. Therefore, the line $y=\frac{1}{1}=1$ is the horizontal asymptote of the function.

Intercepts

To find the $x\text{-}$intercepts, $0$ is substituted for $E(x)$ into the function. It will be better to use the form in which the numerator and the denominator are factored. To find the $y\text{-}$intercept, substitute $0$ for $x$ in the function and solve for $E(0).$ This time the form of the function found in Part A can be used. There is a $y\text{-}$intercept at $(0,\frac{1}{3}).$ Finally, all the intercepts can be plotted on the coordinate plane.

Table of Values

Now, make a table of values to graph the given function.

$x$	$\frac{x^2+4x-2}{x^2+x-6}$	$E(x)=\frac{x^2+4x-2}{x^2+x-6}$
$\text{-}6$	$\frac{(\text{-}6{)}^2+4(\text{-}6)-2}{(\text{-}6{)}^2+(\text{-}6)-6}$	$\approx 0.417$
$\text{-}3.5$	$\frac{(\text{-}3.5{)}^2+4(\text{-}3.5)-2}{(\text{-}3.5{)}^2+(\text{-}3.5)-6}$	$\approx \text{-}1.364$
$\text{-}2$	$\frac{(\text{-}2{)}^2+4(\text{-}2)-2}{(\text{-}2{)}^2+(\text{-}2)-6}$	$1.5$
$\text{-}1$	$\frac{(\text{-}1{)}^2+4(\text{-}1)-2}{(\text{-}1{)}^2+(\text{-}1)-6}$	$\approx 0.833$
$1$	$\frac{(1{)}^2+4(1)-2}{(1{)}^2+(1)-6}$	$\text{-}0.75$
$3$	$\frac{(3{)}^2+4(3)-2}{(3{)}^2+(3)-6}$	$\approx 3.167$
$4$	$\frac{(4{)}^2+4(4)-2}{(4{)}^2+(4)-6}$	$\approx 2.143$
$5$	$\frac{(5{)}^2+4(5)-2}{(5{)}^2+(5)-6}$	$\approx 1.792$

Finally, the graph of the function can be drawn by plotting the found points and connecting them with a smooth curve. Recall that a rational function can cross the horizontal asymptote but cannot cross the vertical asymptotes. In this case, the horizontal asymptote is crossed.