Question

question
find all vertical asymptotes of the following function.
$f(x) = \frac{2x^2 + 17x + 21}{2x^2 + 16x + 14}$
answer attempt 1 out of 2
no vertical asymptotes

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(2x^{2}+17x + 21\). We need two numbers that multiply to \(2\times21 = 42\) and add to \(17\). The numbers are \(14\) and \(3\). So,

$$
LATEXBLOCK0
$$

Next, factor the denominator \(2x^{2}+16x + 14\). Factor out a \(2\) first: \(2(x^{2}+8x + 7)\). Then factor \(x^{2}+8x + 7\), we need two numbers that multiply to \(7\) and add to \(8\), which are \(7\) and \(1\). So,

$$
LATEXBLOCK1
$$

So the function becomes \(f(x)=\frac{(2x + 3)(x + 7)}{2(x + 7)(x + 1)}\)

Step2: Cancel common factors

We can cancel the common factor \((x + 7)\) (assuming \(x
eq - 7\)), so \(f(x)=\frac{2x + 3}{2(x + 1)}\) for \(x
eq - 7\)

Step3: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero (after canceling common factors). Set the denominator of the simplified function equal to zero: \(2(x + 1)=0\)
Solve for \(x\): \(x+1 = 0\Rightarrow x=-1\)
We also need to check the value \(x = - 7\) which was canceled. When \(x=-7\), the original function has a hole (since the factor \((x + 7)\) is canceled), not a vertical asymptote.

Answer:

The vertical asymptote is \(x=-1\)