Based only on the vertical asymptotes, which of the following graphs could be the graph of the given function? $f\left(x\right)=\frac{x^2-4x}{x^2-x-12}$

Graph showing vertical asymptotes at x=-4 and x=3 for the function f(x).

Graph showing a vertical asymptote at x=-3 for the function f(x).

Graph showing vertical asymptotes at x=4 for the function f(x).

Verified step by step guidance

Identify the vertical asymptotes of the function by setting the denominator equal to zero: $x^2 - x - 12 = 0$.

Factor the quadratic equation in the denominator: $x^2 - x - 12 = (x - 4)(x + 3)$.

Set each factor equal to zero to find the values of $x$ that make the denominator zero: $x - 4 = 0$ and $x + 3 = 0$.

Solve these equations to find the vertical asymptotes: $x = 4$ and $x = -3$.

Compare the vertical asymptotes $x = 4$ and $x = -3$ with the graphs provided to determine which graph has these asymptotes.

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Based only on the vertical asymptotes, which of the following graphs could be the graph of the given function? f(x)=x2−4xx2−x−12f\left(x\right)=\frac{x^2-4x}{x^2-x-12}f(x)=x2−x−12x2−4x