Question

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

select one:
a. ( y = \frac{-2(x-2)^2}{(x+1)} )
b. ( y = \frac{(x-2)^2}{2(x+1)} )
c. ( y = \frac{8(x+1)}{(x-2)^2} )
d. ( y = \frac{2(x-2)^2}{(x+1)} )

Explanation:

Step1: Set rational function template

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

Step2: Solve for constant $k$

Use the y-intercept $(0,2)$: substitute $x=0$, $y=2$ into the equation:
$$2 = \frac{k(0-2)^2}{0+1}$$
Simplify the right-hand side:
$$2 = \frac{k(4)}{1}$$
Solve for $k$:
$$k = \frac{2}{4} = \frac{1}{2}$$

Step3: Substitute $k$ into template

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

Answer:

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