Answer
The values of a, b and c in the function $f\left( x \right)=a{{x}^{2}}+bx+c$ are $8,20\text{ and }-12$ respectively.
Work Step by Step
$f\left( x \right)=a{{x}^{2}}+bx+c$ …… (1)
Also, $f\left( -3 \right)=0,f\left( \frac{1}{2} \right)=0,f\left( 0 \right)=-12$implies the ordered pairs $\left( -3,0 \right),\left( \frac{1}{2},0 \right),\left( 0,-12 \right)$.
Now substitute the ordered pairs in equation (1).
For $\left( -3,0 \right)$
$\begin{align}
& 0=a{{\left( -3 \right)}^{2}}-3b+c \\
& =9a-3b+c
\end{align}$ …… (2)
For $\left( \frac{1}{2},0 \right)$
$\begin{align}
& 0=a{{\left( \frac{1}{2} \right)}^{2}}+\left( \frac{1}{2} \right)b+c \\
& =\frac{1}{4}a+\frac{1}{2}b+c
\end{align}$ …… (3)
For $\left( 0,-12 \right)$
$\begin{align}
& -12=a{{\left( 0 \right)}^{2}}+0b+c \\
& -12=c
\end{align}$
Put the value of c in (2),
$\begin{align}
& 0=9a-3b-12 \\
& 12=9a-3b \\
& 4=3a-b
\end{align}$ …… (4)
Put the value of c in (3),
$\begin{align}
& 0=\frac{1}{4}a+\frac{1}{2}b+\left( -12 \right) \\
& 12=\frac{1}{4}a+\frac{1}{2}b \\
& 12=\frac{a+2b}{4} \\
& 48=a+2b
\end{align}$ …… (5)
Now multiply (4) with 2 and add to (5),
$\begin{align}
& \text{ }8=6a-2b \\
& \underline{48=\text{ }a+2b} \\
& 56=7a \\
\end{align}$
Solve it further to get,
$a=8$
Now put the value of a in (4) to get,
$\begin{align}
& 4=3\left( 8 \right)-b \\
& 4=24-b \\
& 24-4=b \\
& 20=b
\end{align}$
The values of a, b and c in the function $f\left( x \right)=a{{x}^{2}}+bx+c$ are $8,20\text{ and }-12$ respectively.
