Question

Find a,b, c when $f\left(x\right)=a{x}^2+bx+c=0$ and $f\left(0\right)=6,f\left(2\right)=11,f(-3)=6$.
Determine $f\left(x\right)$ to find the value of $f\left(1\right)$.

Answer 1

$f\left(0\right)=a\ast 0+b\ast 0+c=6\Rightarrow c=6f(-3)=9a-3b+6=6\Rightarrow 9a=3b\Rightarrow 3a=bf\left(2\right)=4a+2b+6=11\Rightarrow 4a+2b=5\Rightarrow 4a+2\ast 3a=5\Rightarrow a=\frac{1}{2}$

and $b=\frac{3}{2}$

Hence, $f\left(x\right)=\frac{1}{2}{x}^2+\frac{3}{2}x+6\Rightarrow f\left(1\right)=\frac{1}{2}+\frac{3}{2}+6=8$