Thanks for your help guys, this question is still bothering me though. Whoever can solve this is a genius! I can't believe this question was from C1.
The equation of a curve is
![y = ax^2 - 2bx + c](https://studentforums.biz/cgi-bin/mimetex.cgi?y = ax^2 - 2bx + c)
, where
a,
b and
c are constants with
![a > 0](https://studentforums.biz/cgi-bin/mimetex.cgi?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](https://studentforums.biz/cgi-bin/mimetex.cgi?y = x)
, find an expression for
c in terms of
a and
b. Show that in this case, whatever the value of
b,
![c \geq - \frac{1}{4a}](https://studentforums.biz/cgi-bin/mimetex.cgi?c \geq - \frac{1}{4a})
.
I've partially answered the question (I've done parts a) and a bit of b), but don't know how to show that the statement in b) is correct).
Here are my answers so far:
a)
![(\frac{b}{a} , c-\frac{b^2}{a})](https://studentforums.biz/cgi-bin/mimetex.cgi?(\frac{b}{a} , c-\frac{b^2}{a}))
b)
![c = \frac{b(b+1)}{a}](https://studentforums.biz/cgi-bin/mimetex.cgi?c = \frac{b(b+1)}{a})
, but I don't know how to 'show that...'.
Many many thanks in advance!