Question

At which values of x does the graph of the function F(x) have a vertical asymptote? Select all that apply. F(x) = x+4/x^2 +5x -24

Answer 1

To find the vertical asymptotes of the function \( F(x) = \frac{x + 4}{x^2 + 5x - 24} \), we need to determine when the denominator is equal to zero, as vertical asymptotes occur where the function goes to infinity (which happens when the denominator is zero and the numerator is not zero).

First, we set the denominator equal to zero:

\[
x^2 + 5x - 24 = 0
\]

Next, we can factor the quadratic. We need two numbers that multiply to \(-24\) and add up to \(5\). These numbers are \(8\) and \(-3\). Therefore, we can factor the quadratic as:

\[
(x + 8)(x - 3) = 0
\]

Setting each factor to zero gives us the potential vertical asymptotes:

1. \(x + 8 = 0 \Rightarrow x = -8\)
2. \(x - 3 = 0 \Rightarrow x = 3\)

Next, we need to check that the numerator, \(x + 4\), does not equal zero at these values, as the function could have a hole instead of a vertical asymptote if the numerator is also zero.

For \(x = -8\):

\[
x + 4 = -8 + 4 = -4 \quad (\text{not zero})
\]

For \(x = 3\):

\[
x + 4 = 3 + 4 = 7 \quad (\text{not zero})
\]

Since the numerator is not zero for either value, we conclude that the function has vertical asymptotes at both points.

Thus, the values of \(x\) at which the graph of the function \(F(x)\) has vertical asymptotes are:

\(\boxed{-8}\) and \(\boxed{3}\).