Thread: Quadratics

Can you please help me with the last part of this question...

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

So what did you get for parts a) and b)?

Originally Posted by Prove It

So what did you get for parts a) and b)?

For a) I completed the square and got the coordinated as
( b/a , c-(b^2/a) )
b) c-(b^2/a)=b/a
c=b(b-1)/a

I am confused in the last part..
I don't get it why whatever the value of b, c should always be greater than or equal to -1/4a

I had an idea of making b=o and finding the value of c but that won't work i'll get 0