Given that $\displaystyle y = 5x^2 +ax + b $ has a turning point at (b,a). Where a does not equal 0, find a and b.
I know that I need to get two equations to solve simultaneously, but not sure what the equations should look like.
Given that $\displaystyle y = 5x^2 +ax + b $ has a turning point at (b,a). Where a does not equal 0, find a and b.
I know that I need to get two equations to solve simultaneously, but not sure what the equations should look like.
If you don't like calculus, start by completing the square:
$\displaystyle 5\left(x^2+\frac{ax}{5}+~~~\right)+b = 5\left(x^2+\frac{ax}{5}+\frac{a^2}{100}\right)+b-\frac{a^2}{20} = 5\left(x+\frac{a}{10}\right)+\left(b-\frac{a^2}{20}\right)$
So it has a turning point at $\displaystyle \left(-\frac{a}{10}, b-\frac{a^2}{20}\right)$. However, you are given that its turning point is at $\displaystyle (b,a)$
So now you have two equations:
$\displaystyle b=-\frac{a}{10}~~~~~(1)$
$\displaystyle a=b-\frac{a^2}{20}~~~~~(2)$
Spoiler: