Question

Let $p\left(x\right)=a{x}^2+bx+8$ this quadratic polynomial if the min value of the p(x) i.e $6$ when $x=2$ and find the values of $a$ and $b$

Answer 1

$p\left(x\right)=a{x}^2+bx+8$

$p\left(2\right)=6$

$4a+2b+8=6$

$4a+2b=-2$

The minimum value of $p\left(x\right)$ occurs when

$x=\frac{-b}{2a}$

$\therefore 2=\frac{-b}{2a}$

$4a+b=0$

$4a+2b=-2$

$(-)$ $(-)$

________________

$-b=-2$

$b=2$

$4a+2=0$

$a=\frac{-2}{4}=\frac{-1}{2}$

$a=\frac{-1}{2}$

$b=2$.