Each vertical asymptote corresponds to a zero in the denominator. When the function does not change sign from one side of the asymptote to the other, the factor has even degree. The vertical asymptote at x=-4 corresponds to a denominator factor of (x+4). The one at x=2 corresponds to a denominator factor of (x-2)², because the function does not change sign there.

Each zero corresponds to a numerator factor that is zero at that point. Again, if the sign doesn't change either side of that zero, then the factor has even multiplicity. The zero at x=1 corresponds to a numerator factor of (x-1)².

Each "hole" in the function corresponds to numerator and denominator factors that are equal and both zero at that point. The hole at x=-3 corresponds to numerator and denominator factors of (x-3).

Taken altogether, these factors give us the function ...

$\boxed{f(x)=\dfrac{(x+3)(x-1)^2}{(x+4)(x+3)(x-2)^2}}$

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3 years ago