Question

what are the vertical asymptotes of the function $f(x) = \frac{3x + 9}{x^2 + 4x - 12}$?
\bigcirc $x = -6$ and $x = -2$
\bigcirc $x = -6$ and $x = 2$
\bigcirc $x = 1$ and $x = -2$
\bigcirc $x = 1$ and $x = 2$

Explanation:

Step1: Factor the denominator

To find vertical asymptotes, we first factor the denominator \(x^2 + 4x - 12\). We need two numbers that multiply to \(-12\) and add to \(4\). Those numbers are \(6\) and \(-2\). So, \(x^2 + 4x - 12=(x + 6)(x - 2)\).

Step2: Find the values that make the denominator zero

Set the factored denominator equal to zero: \((x + 6)(x - 2)=0\). Using the zero - product property, we have \(x+6 = 0\) or \(x - 2=0\). Solving \(x+6 = 0\) gives \(x=-6\), and solving \(x - 2=0\) gives \(x = 2\). We also need to check that these values do not make the numerator zero. The numerator is \(3x + 9\). For \(x=-6\), \(3(-6)+9=-18 + 9=-9
eq0\). For \(x = 2\), \(3(2)+9=6 + 9 = 15
eq0\). So the vertical asymptotes occur at \(x=-6\) and \(x = 2\).

Answer:

x = -6 and x = 2 (the option with "x = -6 and x = 2")