#### furor celtica

the equation of a curve is y=a(x^2)-2bx+c, where a, b and c are constants with a>0
given that the verte 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>= -1/4a

so i got c= b(b+1)/a or something similar, but the problem is i am totally lost as to where this equation fits in with c>= -1/4a
can someone explain this to me

#### Dinkydoe

Hee Furor,

I assume with verte, you meant 'vertex'.

I can't make pictures, but try to imagine a parabola above the line $$\displaystyle y=x$$.

Since the line $$\displaystyle y=x$$ hits the top of the parabola $$\displaystyle ax^2-2bx+c$$, we must have exactly one root of the equation:

$$\displaystyle ax^2-2bx+c = x$$
or

$$\displaystyle ax^2-(2b+1)x+c=0$$

This happens when the discriminant $$\displaystyle \Delta(ax^2-(2b+1)x+c) = (2b+1)^2-4ac =0$$

Rewriting gives $$\displaystyle 4ac = (2b+1)^2$$ or $$\displaystyle c = \frac{(2b+1)^2}{4a}\geq -\frac{1}{4a}$$ (since a>0 )

the equation of a curve is y=a(x^2)-2bx+c, where a, b and c are constants with a>0
given that the verte 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>= -1/4a

so i got c= b(b+1)/a or something similar, but the problem is i am totally lost as to where this equation fits in with c>= -1/4a
can someone explain this to me
hi

rearrange your final equation to get b^2+b-ac=0

then for all real values of b, 1^2-4(-ac)>=0

which conclude the last part

thanks y'all

#### furor celtica

oh wait dinkydoe i don't get that last part where (2b+1)^2/4a >= -1/4a
i get that the numerator >= 0 cos its a square
and that the denominator > 0 cos a > than zero
but then what?

#### Dinkydoe

Like you said $$\displaystyle (2b+1)^2/4a \geq 0$$ and we have $$\displaystyle -\frac{1}{4a} < 0$$ ( since a > 0 the whole number must be negative)

So naturally we have: $$\displaystyle (2b+1)^2/4a \geq -\frac{1}{4a}$$

I think you'll agree with me that a positive number is bigger then a negative one.

So it's actually a strict inequality '$$\displaystyle >$$'.