Question

find the equation of all vertical asymptotes of the following function. f(x) = \frac{3x + 3}{\sqrt{-x^{2}-x + 20}}

Explanation:

Step1: Set denominator equal to 0

We need to solve $\sqrt{-x^{2}-x + 20}=0$. First, set $-x^{2}-x + 20 = 0$.

Step2: Solve quadratic equation

For the quadratic equation $-x^{2}-x + 20=0$, multiply through by - 1 to get $x^{2}+x - 20=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b=1$ and $c=-20$.
\[

$$\begin{align*} x&=\frac{-1\pm\sqrt{1^{2}-4\times1\times(-20)}}{2\times1}\\ &=\frac{-1\pm\sqrt{1 + 80}}{2}\\ &=\frac{-1\pm\sqrt{81}}{2}\\ &=\frac{-1\pm9}{2} \end{align*}$$

\]
We get $x_1=\frac{-1 + 9}{2}=4$ and $x_2=\frac{-1-9}{2}=-5$.

Step3: Check domain of original function

The function $y = f(x)$ has the condition $-x^{2}-x + 20>0$ (since it's in a square - root in the denominator). The domain of the function is $-x^{2}-x + 20>0$, or $x^{2}+x - 20<0$, which factors to $(x + 5)(x - 4)<0$. The solution of the inequality is $-5

Answer:

$x=-5,x = 4$