Question

write an equation for a rational function with the given characteristics. vertical asymptote at $x = -1$, double zero at $x = 2$, y-intercept at $(0, 2)$.

\\(\circ\\) a. \\(y = \dfrac{x + 1}{2(x - 2)^2}\\)

\\(\circ\\) b. \\(y = \dfrac{(x - 2)^2}{2(x + 1)}\\)

\\(\circ\\) c. \\(y = \dfrac{(x - 2)^2}{4(x + 1)}\\)

\\(\circ\\) d. \\(y = \dfrac{2(x - 2)}{x + 1}\\)

Explanation:

Step1: Set rational function template

A rational function with a double zero at $x=2$ has $(x-2)^2$ in the numerator, and a vertical asymptote at $x=-1$ has $(x+1)$ in the denominator. Let the function be $y=\frac{k(x-2)^2}{x+1}$, where $k$ is a constant.

Step2: Solve for $k$ using y-intercept

Substitute $x=0, y=2$ into the function:
$2=\frac{k(0-2)^2}{0+1}$
Simplify: $2=\frac{k(4)}{1}$
Solve for $k$: $k=\frac{2}{4}=\frac{1}{2}$

Step3: Finalize the function

Substitute $k=\frac{1}{2}$ back into the template:
$y=\frac{\frac{1}{2}(x-2)^2}{x+1}=\frac{(x-2)^2}{2(x+1)}$

Step4: Match with options

Compare the derived function to the given choices.

Answer:

b. $y = \frac{(x - 2)^2}{2(x + 1)}$