Question

write an equation of a rational function with the given characteristics. vertical asymptotes at ( x = 4 ) and ( x = -3 ), ( x )-intercepts at ( (3,0) ) and ( (-2,0) ), ( y )-intercept at ( (0, -4) ).
a. ( y = \frac{-(x - 4)(x + 3)}{8(x - 3)(x + 2)} )
b. ( y = \frac{-4(x - 3)(x + 2)}{(x - 4)(x + 3)} )
c. ( y = \frac{-8(x - 3)(x + 2)}{(x - 4)(x + 3)} )
d. ( y = \frac{8(x - 3)(x + 2)}{(x - 4)(x + 3)} )

Explanation:

Step1: Set rational function skeleton

For vertical asymptotes $x=4$ and $x=-3$, denominator is $(x-4)(x+3)$. For $x$-intercepts $(3,0)$ and $(-2,0)$, numerator is $a(x-3)(x+2)$ where $a$ is a constant. So the function is:
$$y=\frac{a(x-3)(x+2)}{(x-4)(x+3)}$$

Step2: Solve for constant $a$

Use $y$-intercept $(0,-4)$: substitute $x=0$, $y=-4$ into the function:
$$-4=\frac{a(0-3)(0+2)}{(0-4)(0+3)}$$
Simplify the right-hand side:
$$-4=\frac{a(-3)(2)}{(-4)(3)}=\frac{-6a}{-12}=\frac{a}{2}$$
Solve for $a$:
$$a=-4\times2=-8$$

Step3: Finalize the function

Substitute $a=-8$ back into the skeleton:
$$y=\frac{-8(x-3)(x+2)}{(x-4)(x+3)}$$

Answer:

C. $y = \frac{-8(x-3)(x+2)}{(x-4)(x+3)}$