Question

find the vertical asymptotes, if any, of the graph of the following rational function.

h(x) = \frac{x^2 - 25}{x^2 + 3x - 54}

select the correct choice below and, if necessary, fill in the answer box within your choice.

a. the vertical asymptote(s) is/are \square.
(simplify your answer. type an equation. use a comma to separate answers as needed.)

b. there is no vertical asymptote.

Explanation:

Step1: Factor numerator and denominator

Factor \(x^2 - 25\) as \((x - 5)(x + 5)\) (difference of squares: \(a^2 - b^2=(a - b)(a + b)\) with \(a = x\), \(b = 5\)).
Factor \(x^2+3x - 54\): find two numbers that multiply to \(-54\) and add to \(3\). The numbers are \(9\) and \(-6\), so \(x^2+3x - 54=(x + 9)(x - 6)\).
So \(h(x)=\frac{(x - 5)(x + 5)}{(x + 9)(x - 6)}\).

Step2: Find values that make denominator zero (and not numerator)

Set denominator \((x + 9)(x - 6)=0\).
Solve \(x + 9 = 0\) gives \(x=-9\); solve \(x - 6 = 0\) gives \(x = 6\).
Check numerator at \(x=-9\): \((-9 - 5)(-9 + 5)=(-14)(-4)
eq0\).
Check numerator at \(x = 6\): \((6 - 5)(6 + 5)=(1)(11)
eq0\).

Answer:

A. The vertical asymptote(s) is/are \(x=-9, x = 6\)