f(x)=\frac{2x^{2}+10x + 8}{4x^{2}-12x - 16}.\nfind the equations of the horizontal asymptotes and the…

f(x)=\frac{2x^{2}+10x + 8}{4x^{2}-12x - 16}.\nfind the equations of the horizontal asymptotes and the vertical asymptotes of f(x). enter dne if there is no such asymptote. if there is more than one asymptote of a given type, enter them as a comma separated list.\nhorizontal asymptotes:\nvertical asymptotes:

f(x)=\frac{2x^{2}+10x + 8}{4x^{2}-12x - 16}.\nfind the equations of the horizontal asymptotes and the vertical asymptotes of f(x). enter dne if there is no such asymptote. if there is more than one asymptote of a given type, enter them as a comma separated list.\nhorizontal asymptotes:\nvertical asymptotes:

Answer

Explanation:

Step1: Simplify the rational function

First, factor the numerator and denominator. The numerator $2x^{2}+10x + 8=2(x^{2}+5x + 4)=2(x + 1)(x + 4)$. The denominator $4x^{2}-12x-16 = 4(x^{2}-3x - 4)=4(x + 1)(x-4)$. So, $f(x)=\frac{2(x + 1)(x + 4)}{4(x + 1)(x - 4)}=\frac{x + 4}{2(x - 4)}$ for $x\neq - 1$.

Step2: Find horizontal asymptotes

For a rational function $y=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots+b_0}$, if $n = m$, the horizontal - asymptote is $y=\frac{a_n}{b_m}$. Here, $n = m=1$, $a_1 = 1$, $b_1 = 2$. So the horizontal asymptote is $y=\frac{1}{2}$.

Step3: Find vertical asymptotes

Vertical asymptotes occur at the values of $x$ that make the denominator equal to zero. Set the denominator $2(x - 4)=0$. Solving for $x$, we get $x = 4$.

Answer:

Horizontal asymptotes: $y=\frac{1}{2}$ Vertical asymptotes: $x = 4$