SOLUTION: The parabola has a turning point at (z, -8). It intersects the y-axis at y=10 and one of the x-intercepts is x=5. Find: the value of z

Algebra ->  Testmodule -> SOLUTION: The parabola has a turning point at (z, -8). It intersects the y-axis at y=10 and one of the x-intercepts is x=5. Find: the value of z      Log On


   

Question 1037252: The parabola has a turning point at (z, -8). It intersects the y-axis at y=10 and one of the x-intercepts is x=5. Find:
the value of z
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Let the quadratic function be f%28x%29+=+a%28x-r%29%28x-5%29, where x=5 is one of the roots (as given via the x-intercept), and x = r is the other root.
Since y = 10 is the value of the y-intercept, the constant term in the quadratic function must be equal to 5ar, and moreover, 5ar = 10, by hypothesis.
==> ar = 2.
Since the turning point (or the vertex of the parabola) is always midway between the two roots, it follows that z+=+%28r%2B5%29%2F2.
==> , or
a%28r-5%29%5E2+=+32.
Divide this equation by the equation ar = 2, giving
%28r-5%29%5E2%2Fr+=+16, or,
%28r-5%29%5E2+=+16r or
r%5E2-26r%2B25+=+%28r-25%29%28r-1%29+=+0, after expanding and simplifying.
==> r = 25 or r = 1.
If r = 25, then a = 2/25.
If r = 1, then a = 2.
Hence there are two possible answers:
(i) The function f%28x%29+=+%282%2F25%29%28x-25%29%28x-5%29 has z+=+%2825%2B5%29%2F2+=+15, while,
(ii) The function f%28x%29+=+2%28x-1%29%28x-5%29 has z+=+%281%2B5%29%2F2+=+3.