Question

find the equation of all vertical asymptotes of the following function.
f(x)=\frac{2x + 2}{sqrt{-x^{2}+2x + 48}}

Explanation:

Step1: Find when the denominator is zero.

Set $-x^{2}+2x + 48=0$.

Step2: Solve the quadratic equation.

For a quadratic equation $ax^{2}+bx + c = 0$ ($a=-1$, $b = 2$, $c = 48$), use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(2)^{2}-4\times(-1)\times48=4 + 192=196$. Then $x=\frac{-2\pm\sqrt{196}}{-2}=\frac{-2\pm14}{-2}$. We get $x_1=\frac{-2 + 14}{-2}=-6$ and $x_2=\frac{-2-14}{-2}=8$. But we have a square - root in the denominator, and for the function to be well - defined, $-x^{2}+2x + 48>0$. The domain of the function is $-6

Answer:

No Vertical Asymptotes