Question

question
find all vertical asymptotes of the following function.
$f(x) = \frac{2x^2 + 7x}{3x^2 + 3x - 90}$

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(2x^{2}+7x\). We can factor out an \(x\) from both terms:
\(2x^{2}+7x = x(2x + 7)\)

Next, factor the denominator \(3x^{2}+3x - 90\). First, factor out a common factor of 3:
\(3x^{2}+3x - 90=3(x^{2}+x - 30)\)

Then, factor the quadratic inside the parentheses. We need two numbers that multiply to \(- 30\) and add to \(1\). Those numbers are \(6\) and \(-5\). So,
\(x^{2}+x - 30=(x + 6)(x - 5)\)

So the denominator factored completely is \(3(x + 6)(x - 5)\) and the numerator is \(x(2x + 7)\).

Step2: Find values that make denominator zero (excluding holes)

To find vertical asymptotes, we set the denominator equal to zero (after ensuring there are no common factors with the numerator, which there aren't here since the numerator factors are \(x\) and \(2x + 7\) and the denominator factors are \(3\), \(x + 6\), and \(x - 5\)).

Set \(3(x + 6)(x - 5)=0\). Since \(3
eq0\), we set each of the other factors equal to zero:

For \(x+6 = 0\), we solve for \(x\) and get \(x=-6\).

For \(x - 5=0\), we solve for \(x\) and get \(x = 5\).

Answer:

The vertical asymptotes are \(x=-6\) and \(x = 5\)