Question

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

Answer 1

Given $f\left(x\right)=a{x}^2+bx+c$
$\therefore f\left(0\right)=a{.0}^2+b.0+c=6$
or $0.a+0.b+1.c=6$ (i)
and $f\left(2\right)=a(2{)}^2+b(2)+c=11$
$\Rightarrow 4.a+2.b+1.c=11$ (ii)
and $f(-3)=a(-3{)}^2+b(-3)+c=6$
$\Rightarrow 9.a-3.b+c=6$ (iii)
Since (i), (ii) and (iii) are three equations in a, b, c solving these by Cramer's rule
$\therefore D=\left|\begin{array}{ccc}0& 0& 1\\ 4& 2& 1\\ 9& -3& 1\end{array}\right|=1.(-12-18)=-30$
$D_1=\left|\begin{array}{ccc}6& 0& 1\\ 11& 2& 1\\ 6& -3& 1\end{array}\right|=6(2+3)-0+1(-33-12)$ $=-15$
$D_2=\left|\begin{array}{ccc}0& 6& 1\\ 4& 11& 1\\ 9& 6& 1\end{array}\right|$ $=0-6(4-9)+1(24-99)$ $=20-75$ $=-45$
$D_3=\left|\begin{array}{ccc}0& 0& 6\\ 4& 2& 11\\ 9& -3& 6\end{array}\right|$ $=6(-12-18)$ $=-180$
Hence by Cramer's rule
$a=\frac{D_1}{D}=\frac{-15}{-30}=\frac{1}{2}$
$b=\frac{D_2}{D}=\frac{-45}{-30}=\frac{3}{2}$
and $c=\frac{D_3}{D}=\frac{-180}{-30}=6$
Hence $a=\frac{1}{2},b=\frac{3}{2},c=6$
$\therefore f\left(x\right)=\frac{1}{2}{x}^2+\frac{3}{2}x+6$
and $f\left(1\right)=\frac{1}{2}(1{)}^2+\frac{3}{2}(1)+6$ $=8$