Suppose that $f(x)=\frac{4 x^{2}-32 x+60}{x^{2}-11 x+30}$. The function can be written in factored form as $f(x)=\frac{4(x-3)(x-5)}{(x-5)(x-6)}$ a. Provide the location of any removable discontinuities. \[ x= \] Preview b. Provide the location of any vertical asymptotes. \[ x= \] Preview c. As $x \rightarrow \pm \infty, f(x)$ behaves like: \[ y= \] Preview

Step 1 ：The function can be written in factored form as \(f(x)=\frac{4(x-3)(x-5)}{(x-5)(x-6)}\)

Step 2 ：A removable discontinuity occurs when a factor in the denominator of a rational function also appears in the numerator. In this case, the factor \((x-5)\) appears in both the numerator and the denominator, so \(x=5\) is a removable discontinuity.

Step 3 ：A vertical asymptote occurs when a factor in the denominator does not appear in the numerator. In this case, the factor \((x-6)\) only appears in the denominator, so \(x=6\) is a vertical asymptote.

Step 4 ：As \(x\) approaches positive or negative infinity, the function behaves like the ratio of the leading coefficients of the numerator and the denominator. In this case, the leading coefficients are 4 and 1, so the function behaves like \(y=4\).

Step 5 ：The location of the removable discontinuity is \(\boxed{5}\).

Step 6 ：The location of the vertical asymptote is \(\boxed{6}\).

Step 7 ：As \(x \rightarrow \pm \infty, f(x)\) behaves like \(\boxed{4}\).