Given $f ( x ) = \frac { 5 x + 4 } { 5 x ^ { 2 } + 4 x }$. Determine the domain of $f ( x )$ and find any vertical asymptotes.
A) domain: all real numbers except $x = - \frac { 4 } { 5 }$ vertical asymptote: $x = 0$
B) domain: all real numbers except $x = 0$ and $x = - \frac { 4 } { 5 }$ vertical asymptote: $x = 0$
C) domain: all real numbers vertical asymptotes: $x = 0$ and $x = - \frac { 4 } { 5 }$
D) domain: all real numbers except $x = 0$ and $x = \frac { 4 } { 5 }$ vertical asymptote: $x = 0$
E) domain: all real numbers except $x = \frac { 4 } { 5 }$ vertical asymptotes: $x = 0$ and $x = - \frac { 4 } { 5 }$

Question

Quizplus · Accepted Answer

The domain of the function is all real numbers except where the denominator equals zero. Setting the denominator to zero, we get $5x^2+4x=0$, or $x(5x+4)=0$. This gives us $x=0$ and $x=-\frac{4}{5}$ as the values where the denominator equals zero. Thus, the domain of the function is all real numbers except $0$ and $-\frac{4}{5}$. To find the vertical asymptotes, we look for values in the domain where the function becomes unbounded. This occurs when the denominator is very close to zero, but not quite zero. As $x$ gets very large in either direction, the function will approach $0$, so there is no vertical asymptote there. However, as we approach $x=0$ or $x=-\frac{4}{5}$ from either side, the denominator approaches zero while the numerator remains bounded. Thus, we have vertical asymptotes at $x=0$ and $x=-\frac{4}{5}$. The only choice that matches this description is $\boxed{	extbf{(B)}}$.

Given f(x)=5x+45x2+4xf ( x ) = \frac { 5 x + 4 } { 5 x ^ { 2 } + 4 x }f(x)=5x2+4x5x+4​

Given $f ( x ) = \frac { 5 x + 4 } { 5 x ^ { 2 } + 4 x }$