Solve the problem.
-Determine which rational function $\mathrm { R } ( \mathrm { x } )$ has a graph that crosses the $x$-axis at $- 1$, touches the $x$-axis at $- 4$, has vertical asymptotes at $x = - 2$ and $x = 3$, and has one horizontal asymptote at $y = - 2$.
A) $R ( x ) = \frac { - 2 ( x + 1 ) ( x + 4 ) ^ { 2 } } { ( x + 2 ) ^ { 2 } ( x - 3 ) } , x \neq z - 2,3$
B) $R ( x ) = \frac { - 2 ( x + 1 ) ( x + 4 ) } { ( x + 2 ) ( x - 3 ) } , x \neq - 2,3$
C) $R ( x ) = \frac { - ( x + 1 ) ( x + 4 ) ^ { 2 } } { 2 ( x - 2 ) ^ { 2 } ( x + 3 ) } , x \neq 2 , - 3$
D) $R ( x ) = \frac { - 2 ( x - 3 ) ( x + 2 ) ^ { 2 } } { ( x + 4 ) ^ { 2 } ( x + 1 ) } , x \neq - 4 , - 1$

Question

Quizplus · Accepted Answer

The correct function must have factors that correspond to its zeros and asymptotes. For $x$-axis crossings, $x+1$ and $(x+4)^2$ are needed, with the square on $(x+4)$ indicating the touch point at $-4$. Vertical asymptotes at $x=-2$ and $x=3$ require these factors in the denominator, and the leading coefficient and highest power of $x$ in the numerator and denominator determine the horizontal asymptote at $y=-2$, making option A the correct choice.

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