Determine the vertical asymptote(s) of the graph of the function.
-$$h ( x ) = \frac { x + 4 } { 2 x ^ { 2 } + 11 x + 5 }$$
A) $x = - \frac { 1 } { 2 }$ and $x = - 5$
B) $x = \frac { 1 } { 2 }$ and $x = - 5$
C) $x = \frac { 1 } { 2 }$ and $x = - \frac { 5 } { 2 }$
D) $x = 4$ and $x = 5$

Question

Quizplus · Accepted Answer

Vertical asymptotes occur where the denominator of a rational function is zero (assuming the numerator is not zero at the same point, which would indicate a hole rather than an asymptote). For the given function $h(x) = \frac{x + 4}{2x^2 + 11x + 5}$, we set the denominator equal to zero to find the vertical asymptotes: $2x^2 + 11x + 5 = 0$. Factoring or using the quadratic formula, we find the roots of the denominator to be $x = -\frac{1}{2}$ and $x = -5$, which are the locations of the vertical asymptotes.

Determine the Vertical Asymptote(s) of the Graph of the Function h(x)=x+42x2+11x+5h ( x ) = \frac { x + 4 } { 2 x ^ { 2 } + 11 x + 5 }h(x)=2x2+11x+5x+4​

Determine the Vertical Asymptote(s) of the Graph of the Function $h ( x ) = \frac { x + 4 } { 2 x ^ { 2 } + 11 x + 5 }$