#### Answer

$\color{blue}{f(x)=3x^2-15x-18}$

#### Work Step by Step

RECALL:
The zeros of the quadratic function $f(x) =c(x-a)(x-b)=0$ are $x=a$ and $x=b$.
The given quadratic function has the zeros $x=6$ and $x=-1$.
Using the rule mentioned in the recall part above, then the function is:
$f(x)=c(x-6)[x-(-1))
\\f(x)=c(x-6)(x+1)
\\f(x)=c[x(x)+x(1) - 6(x) - 6(1)]
\\f(x)=c(x^2+x-6x-6)
\\f(x)=c(x^2-5x-6)$
With $f(2) = -36$, substitute $-36$ to $f(x)$ and $2$ to $x$ to obtain:
$f(2) = c(2^2-5\cdot 2-6)
\\-36 = c(4-10-6)
\\-36=c(-12)
\\\frac{-36}{-12}=c
\\3=c$
Thus, the function is:
$f(x) = 3(x^2-5x-6)
\\\color{blue}{f(x)=3x^2-15x-18}$