Question

Which of the following functions has a vertical asymptote at x=3, a horizontal asymptote at f(x)=2, and a root at x=5?

Answer 1

To have a vertical asymptote, is the value in which the denominator is zero, therefore if vertical asymptote x=3, then the denominator must be x-3.

Then we find the root, which is the x-intercepts, let's find them:

$\begin{gathered} \frac{-4}{x-3}+2=0 \\ \text{ }\frac{-4}{x\text{ - 3}}=\text{ - 2} \\ \text{ -4 = - 2\lparen x - 3\rparen} \\ \text{ }\frac{-4}{\text{ -2}}=\text{ x - 3 } \\ \\ 2=x\text{ - 3} \\ x=5 \end{gathered}$

So, the function D has the requirements