quadratics

May 2009
271
7
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
 
Dec 2009
411
131
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 )
 
Sep 2008
1,261
539
West Malaysia
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
 
May 2009
271
7
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?
 
Dec 2009
411
131
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 '\(\displaystyle >\)'.