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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>-1/4a or c=-1/4a.
So the vertex is on (b/a, c-b^2/a) and c=b(b+1)/a. Any ideas about the last part?

falzoon · Accepted Answer

y = ax^2 - 2bx + c

Complete the square :

y =a[x^2 - (2b/a)x + c/a]

y = a[x^2 - (2b/a)x + b^2/a^2 + c/a - b^2/a^2]

y = a[(x - b/a)^2 + c/a - b^2/a^2]

y = a(x - b/a)^2 + c - b^2/a

This is now in vertex form, y = a(x - h)^2 + k, where vertex is (h, k),
so, vertex is at (b/a, c - b^2/a).

If vertex lies on y = x, then with x = b/a, we also have y = b/a.
Plugging these into the vertex equation gives :
b/a = a(b/a - b/a)^2 + c - b^2/a

b/a = c - b^2/a

c = b/a + b^2/a = b(b + 1)/a


Now, b(b + 1)/a = c
so, b(b + 1) = ac
or, b^2 + b - ac = 0
Solving for b using the quadratic formula gives :
b = [-1 ± √(1 + 4ac)]/2

For b to be real, term under the square root sign must be ≥ 0,
that is, 1 + 4ac ≥ 0, or c ≥ -1/(4a).

Who can help me solve this exercise?

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