Question

question
find all vertical asymptotes of the following function.
$f(x) = \frac{2x^2 - 32}{5x^2 + 23x + 12}$

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(2x^{2}-32\). We can factor out a 2: \(2(x^{2} - 16)\), and then use the difference of squares \(a^{2}-b^{2}=(a - b)(a + b)\), so \(2(x - 4)(x + 4)\).

Next, factor the denominator \(5x^{2}+23x + 12\). We need two numbers that multiply to \(5\times12 = 60\) and add to 23. Those numbers are 20 and 3. So we rewrite the middle term: \(5x^{2}+20x+3x + 12\). Then factor by grouping: \(5x(x + 4)+3(x + 4)=(5x + 3)(x + 4)\).

So the function becomes \(f(x)=\frac{2(x - 4)(x + 4)}{(5x + 3)(x + 4)}\).

Step2: Cancel common factors and find restrictions

We can cancel the common factor \((x + 4)\) (note that \(x
eq - 4\) to avoid division by zero in the original function). So the simplified function is \(f(x)=\frac{2(x - 4)}{5x+3}\), with \(x
eq - 4\) and \(x
eq-\frac{3}{5}\) (since \(5x+3 = 0\) when \(x=-\frac{3}{5}\)).

Vertical asymptotes occur where the denominator is zero and the numerator is not zero (after canceling common factors). For the simplified denominator \(5x + 3\), setting \(5x+3=0\) gives \(x=-\frac{3}{5}\). We also need to check the canceled factor: when \(x=-4\), the original denominator is zero (since \((5(-4)+3)(-4 + 4)=( - 20 + 3)(0)=(-17)(0)=0\)) but the numerator is \(2(-4 - 4)(-4 + 4)=2(-8)(0)=0\), so \(x = - 4\) is a hole (a removable discontinuity), not a vertical asymptote.

Answer:

The vertical asymptote is \(x =-\frac{3}{5}\)