Question

Write a rational function $g$ with a vertical asymptotes at $x=3$ and $x=-3$, a horizontal asymptote at $y=-4$ and with no $x$ intercept.

• A. $f\left(x\right)=\frac{[4x^2+6]}{\left[\right(x-3\left)\right(x+3\left)\right]}$
• B. $f\left(x\right)=\frac{[-4x^2-6]}{\left[\right(x-3\left)\right(x+3\left)\right]}$
• C. $f\left(x\right)=\frac{[4x^2-6]}{\left[\right(x-3\left)\right(x+3\left)\right]}$
• D. $f\left(x\right)=\frac{[-4x^2+6]}{\left[\right(x-3\left)\right(x+3\left)\right]}$

Answer 1

The correct option is B $f\left(x\right)=\frac{[-4x^2-6]}{\left[\right(x-3\left)\right(x+3\left)\right]}$

Since $g$ has a vertical is at $x=3$ and $x=-3$, then the denominator of the rational function contains the product of $(x-3)$ and $(x+3)$. Function $g$ has the form.

$g\left(x\right)=h\left(x\right)/\left[\right(x-3\left)\right(x+3\left)\right]$

For the horizontal asymptote to exist, the numerator $h\left(x\right)$ of $g\left(x\right)$ has to be of the same degree as the denominator with a leading coefficient equal to $-4$. At the same time $h\left(x\right)$ has no real zeros. Hence

$f\left(x\right)=[-4x^2-6]/\left[\right(x-3\left)\right(x+3)$