1. ## turning point, coordinates

Given that $y = 5x^2 +ax + b$ 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.

2. $y'=10x+a=0$
$y'(b)=10b+a=0$ (1)

$y=a=5b^2+ab+b$ (2)

3. Originally Posted by Tweety
Given that $y = 5x^2 +ax + b$ 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:

$5\left(x^2+\frac{ax}{5}+~~~\right)+b = 5\left(x^2+\frac{ax}{5}+\frac{a^2}{100}\right)+b-\frac{a^2}{20} = 5\left(x+\frac{a}{10}\right)+\left(b-\frac{a^2}{20}\right)$

So it has a turning point at $\left(-\frac{a}{10}, b-\frac{a^2}{20}\right)$. However, you are given that its turning point is at $(b,a)$

So now you have two equations:

$b=-\frac{a}{10}~~~~~(1)$
$a=b-\frac{a^2}{20}~~~~~(2)$

Spoiler:
Subbing into the second equation, we have $a=-\frac{a}{10}-\frac{a^2}{20} \implies a = \frac{-2a-a^2}{20} \implies 20a=-2a-a^2 \implies a^2+22a = 0$

Solving this equation gives us $a=-22$ (it can't be 0 from you initial condition) and it follows from (1) that $b=-\frac{11}{5}$.

So your equation is $\boxed{5x^2-22x-\frac{11}{5}}$