Let P(x) = ax² +bx+ 8 is a quadratic polynomial. If the minimum value of P(x) is 6 at x = 2 then find the value of a and b.

We have .f{x) = ax² +bx+ 8

$\begin{array}{r}=a\left\{{\left(x+\frac{b}{2a}\right)}^2+\frac{8}{a}-\frac{b^2}{4a^2}\right\}\\ =a\left\{{\left(x+\frac{b}{2a}\right)}^2-\frac{b^2-32a}{4a^2}\right\}\end{array}$

Clearly, $\frac{b}{2a}=-2$

$\begin{array}{l}b=-4a\\ \frac{32a-b^2}{4a}=6\\ \frac{32a-16a^2}{4a}=6\\ 8a-4a^2=6a\\ 4a^2=2a\\ a=\frac{1}{2}\end{array}$

b =-2