Numeric 1 point Given f(x)=(x-5)/(2(x+8)(x-5)) , where is the vertical asymptote? x= square

To find the vertical asymptote of the function $f(x)=\frac {x-5}{2(x+8)(x-5)}$, we need to determine the values of $x$ for which the denominator is equal to zero.<br /><br />The denominator of the function is $2(x+8)(x-5)$. To find the values of $x$ that make the denominator zero, we set the denominator equal to zero and solve for $x$:<br /><br />$2(x+8)(x-5) = 0$<br /><br />$(x+8)(x-5) = 0$<br /><br />This equation is satisfied when either $(x+8) = 0$ or $(x-5) = 0$. Solving these equations, we get:<br /><br />$x+8 = 0 \Rightarrow x = -8$<br /><br />$x-5 = 0 \Rightarrow x = 5$<br /><br />Therefore, the vertical asymptotes of the function $f(x)$ are at $x = -8$ and $x = 5$.