Question

3 select the correct answer. which statement describes the graph of the function $h(x)=\frac{4x^{2}-100}{8x - 20}$? a. there is a horizontal asymptote at $y=\frac{1}{2}$. b. there is a horizontal asymptote at $y = 5$. c. the graph has an oblique asymptote. d. there is a horizontal asymptote at $y = 1$.

Explanation:

Step1: Simplify the function

First, factor the numerator and denominator.
$h(x)=\frac{4x^{2}-100}{8x - 20}=\frac{4(x^{2}-25)}{4(2x - 5)}=\frac{(x + 5)(x - 5)}{2x-5}$
Since the degree of the numerator ($n = 2$) is one more than the degree of the denominator ($m=1$) (where $n$ is the degree of the polynomial in the numerator and $m$ is the degree of the polynomial in the denominator).

Step2: Determine the asymptote type

When $n=m + 1$ for a rational - function $y=\frac{f(x)}{g(x)}$, the graph has an oblique asymptote. We can find the oblique asymptote by using polynomial long - division. Divide $x^{2}-25$ by $2x - 5$.
Dividing $x^{2}$ by $2x$ gives $\frac{1}{2}x$. Multiply $2x - 5$ by $\frac{1}{2}x$ to get $x^{2}-\frac{5}{2}x$. Subtract this from $x^{2}-25$ gives $\frac{5}{2}x-25$. Dividing $\frac{5}{2}x$ by $2x$ gives $\frac{5}{4}$.
$x^{2}-25=(2x - 5)(\frac{1}{2}x+\frac{5}{4})-\frac{75}{4}$
As $x\to\pm\infty$, the function $h(x)$ behaves like the line $y=\frac{1}{2}x+\frac{5}{4}$, which is an oblique asymptote.

Answer:

C. The graph has an oblique asymptote.