Question

find equations for the vertical asymptotes, if any, for the following rational function,
$f(x) = \frac{12x^2 + 15x + 3}{-3x - 6}$
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Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(12x^2 + 15x + 3\). We can factor out a 3 first: \(3(4x^2 + 5x + 1)\). Then factor the quadratic: \(3(4x + 1)(x + 1)\).

Now factor the denominator \(-3x - 6\). Factor out a -3: \(-3(x + 2)\).

So the function becomes \(f(x)=\frac{3(4x + 1)(x + 1)}{-3(x + 2)}\). We can simplify the constants: \(\frac{3}{-3}=-1\), so \(f(x)=-\frac{(4x + 1)(x + 1)}{x + 2}\) (note that we have to check for any common factors that would cancel, but here there are no common factors between numerator and denominator).

Step2: Find vertical asymptotes

Vertical asymptotes of a rational function occur where the denominator is zero (and the numerator is not zero at that point, since if numerator and denominator are zero there, it's a hole instead).

Set the denominator equal to zero: \(x + 2 = 0\). Solving for \(x\), we get \(x=-2\).

Now check the numerator at \(x = -2\): substitute \(x=-2\) into the numerator \(-(4x + 1)(x + 1)\). We get \(-(4(-2)+1)(-2 + 1)=-( -8 + 1)(-1)=-(-7)(-1)=-7\), which is not zero. So there is a vertical asymptote at \(x=-2\).

Answer:

\(x = -2\)