Question 214367: Can anyone help me with this problem? Find b and c so that the parabola y = -14 x^2 + b x + c has vertex (6 , 0 ). Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! The vertex of a parabola is at x = -b/2a
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your equation is:
-14x^2 + bx + c = y
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This equation is in standard form when y = 0.
We get:
-14x^2 + bx + c = 0
a = -14
b and c are unknown.
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Equation for the vertex is x = -b/2a
We know that we want the vertex at (6,0), so we are at the vertex when:
x = 6
y = 0
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substituting x = 6 in the equation for the vertex, we get:
6 = -b/2a
We know that a = -14, so this equation becomes:
6 = -b/(-28)
We solve for b to get:
b = 168
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Our equation now becomes:
-14x^2 + 168x + c = y
We know that the vertex is at (6,0) so we know that we want this equation to equal to 0 at the point x = 6.
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We set y = 0 and we substitute 6 for x in the equation to get:
-14*(6^2) + 168(6) + c = 0
This becomes:
-504 + 1008 + c = 0
Thie becomes:
504 + c = 0
we solve for c to get:
c = -504
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Our equation becomes:
y = -14x^2 + 168x - 504
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Answer to your question is:
b = 168
c = -504
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Graph of this equation is shown below:
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Because a is negative, this graph opens downward.
If a was positive, it would open upward.
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