# Calculus

I need help with this question...

Find the values for a, b, and c such that the graph of y= ax^2 + bx + c has a relative maximum at (3,12) and crosses the y-axis at (0,1).

If someone could help me out, that would be awesome :)

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1. V(3,12), (0,1).

Y = a(x - 3)^2 + 12,
Y = a(0-3)^2 + 12 = 1,
9a + 12 = 1,
9a = 1 - 12 = -11,
a = -11/9. Since a is neg, the parabola
opens downward.

Y = -11/9(x-3)^2 + 12,
Y = -11/9(X^2 -6X + 9) + 12,
Y = -11X^2/9 + 22X/3 + 1,

a = -(11/9).

b = 22/3.

c = 1.

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