Question

Need help with number 5. Verify the holes, vertical asymptotes, and horizontal asymptotes of each. Then sketch the graph

Answer 1

Function given:

$f\mleft(x\mright)=\frac{1}{2x^2+2x-12}$

Procedure

• Finding the hole

Factor the polynomial in the denominator:

$f\mleft(x\mright)=\frac{1}{2(x^2+x-6)}$$f(x)=\frac{1}{2\lbrack(x+3)(x-2)\rbrack}$

If there was a hole, there would have to be a common factor between numerator and denominator. Therefore, as this function has no common factor between numerator and denominator, then there is no hole for the rational function.

• Vertical asymptotes

In this case, we also have to factor the denominator. As we already did, we know that the factored expression is:

$f(x)=\frac{1}{2\lbrack(x+3)(x-2)\rbrack}$

Then, we have to equal those factors to 0, meaning:

$x+3=0$$x-2=0$

Solving for x, we get the vertical asymptotes:

$x=-3$$x=2$

Then, the vertical asymtotes are x = -3 and x = 2.

• Horizontal asymptotes

In this case, we have to calculate the limits of the function:

$\lim _{x\to\infty}(\frac{1}{2x^2+2x-12})$$\lim _{x\to-\infty}(\frac{1}{2x^2+2x-12})$

Solving the limits we get:

$\frac{1}{2\infty^2+2\infty-12}=\frac{1}{\infty}=0$$\frac{1}{2(-\infty)^2+2(-\infty)-12}=\frac{1}{\infty}=0$

Then, the horizontal asymptote is y = 0.

Answer:

• Holes: None.

,

• Vertical asymptotes: ,x = -3 ,and ,x = 2,.

,

• Horizontal asymptotes: ,y = 0.

• Graph