#### Answer

$\color{blue}{f(x)=-3x^2-\frac{21}{4}x+\frac{3}{2}}$

#### 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 $x=\frac{1}{4}$ and $x=-2$.
Using the rule mentioned in the recall part above, then the function is:
$f(x)=c(x-\frac{1}{4})[x-(-2)]
\\f(x)=c(x-\frac{1}{4})(x+2)
\\f(x)=c[x(x)+x(2) - \frac{1}{4}(x) - \frac{1}{4}(2)]
\\f(x)=c(x^2+2x-\frac{1}{4}x-\frac{1}{2})
\\f(x)=c(x^2+\frac{7}{4}x-\frac{1}{2})$
With $f(2) = -21$, substitute $-21$ to $f(2)$ and $2$ to $x$ to obtain:
$f(2) = c(2^2+\frac{7}{4}\cdot 2-\frac{1}{2})
\\-21 = c(4+\frac{7}{2}-\frac{1}{2})
\\-21 = c(\frac{8}{2}+\frac{6}{2})
\\-21=c(\frac{14}{2})
\\-21=c(7)
\\\frac{-21}{7}=c
\\-3=c$
Thus, the function is:
$f(x) = -3(x^2+\frac{7}{4}x-\frac{1}{2})
\\\color{blue}{f(x)=-3x^2-\frac{21}{4}x+\frac{3}{2}}$