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Posted by on Tuesday, September 21, 2010 at 2:47pm.

Find expressions for the quadratic functions whose graphs are shown. One graph has the point (4,2) plotted in which the parabola passes through (U-shaped parabola- right side up) The vertex is at (3,0) and the parabola does not touch the y-axis for as much is shown.

So, I did this:

a(3)^2 + b(3) + c = 0
Solve: c = -12

For the point (4,2):
a(4)^2 + b(4) - 12 = 2
Solve: 16a + 4b = 14

I am not sure if I am doing this correctly and I am also stuck at the is point. Please help.

  • Calculus - , Tuesday, September 21, 2010 at 5:19pm

    c does not equal -12:
    a(3)^2 + b(3) + c = 0
    Solve: c = -9a - 3b (not -12)

    For the given parabola, since the vertex is (3,0), and knowing that parabolas are symmetrical with respect to the vertex, we therefore have another point (2,2), the mirror image of (4,2) about x=3 passing through the vertex.

    The three known points give rise to three equations with three unknowns (a,b,c). Solve for a,b and c.

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