#### Answer

$\color{blue}{f(x)=\frac{1}{4}x^2-\frac{3}{4}x-45}$

#### 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 $x=15$ and $x=-12$.
Using the rule mentioned in the recall part above, then the function is:
$f(x)=c(x-15)[x-(-12)]
\\f(x)=c(x-15)(x+12)
\\f(x)=c[x(x)+x(12) - 15(x) - 15(12)]
\\f(x)=c(x^2+12x-15x-180)
\\f(x)=c(x^2-3x-180)$
With $f(3) = -45$, substitute $-45$ to $f(3)$ and $3$ to $x$ to obtain:
$f(3) = c(3^2-3\cdot 3-180)
\\-45 = c(9-9-180)
\\-45=c(-180)
\\\frac{-45}{-180}=c
\\\frac{1}{4}=c$
Thus, the function is:
$f(x) = \frac{1}{4}(x^2-3x-180)
\\\color{blue}{f(x)=\frac{1}{4}x^2-\frac{3}{4}x-45}$