I've been trying to figure out how to do the last part of this question for a long time but I still can't..

The equation of a curve is y=ax^2 -2bx +c ,where a,b and c are constants with a>0
a) Find in terms of a,b and c the coordinates of the vertex of the curve
b) Given that the vertex of the curve lies on the line y=x , find an expression for c in terms of a and b.
Show that in this case, whatever the value of b, c is greater than or equal to -1/4a

For a) I got ( b/a , c - b^2/a )
For b) I got c= (b^2+b) /a

Any help would be greatly appreciated...

2. ## Re: question on quadratics

Hey puresoul.

Can you show us your working out? We can get a better idea of if the answer is right and if not what is going wrong.

3. ## Re: question on quadratics

Originally Posted by chiro
Hey puresoul.

Can you show us your working out? We can get a better idea of if the answer is right and if not what is going wrong.
I thought of this...

We know the vertex of the parabola and the y intercept and since a is greater than zero so c has to be greater than c - (b^2)/a
and c= (b^2 +b) /a
so (b^2 +b) /a >c - (b^2)/a
(2b^2 + b )/a > c
but that doesn't look anything like what they want ..

And I don't know what else I could do
Any hint that might help me?