Question 364006: Create the quadratic equation in the form y=ax^2+bx+c using the point (-5,6) as a point on the graph and (9,-9) as the vertex.
I know the answers, I'm just not sure how to start solving for it.
Found 2 solutions by stanbon, robertb:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Create the quadratic equation in the form y=ax^2+bx+c using the point (-5,6) as a point on the graph and (9,-9) as the vertex.
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Vertex occurs at x = -b/(2a)
Solve 9 = -b/(2a)
-b = 18a
b = -18a
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Substitute into the form to get:
y = ax^2-18ax+c
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Using (-5,6) you get:
6 = 25a+90a+c
6 = 115a + c
Using (9,-9) you get:
-9 = 81a - 162a + c
-9 = -81a + c
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Solve the system:
115a + c = 6
-81a + c = -9
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Subtract and solve for "a":
196a = 15
a = 15/196
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Substitute into 115a + c = 6 to solve for "c":
115(15/196)+c = 6
c = -549/196
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b = -18a = -270/196
Equation:
y = (15/196)x^2 - (270/196)x - (549/196)
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Cheers,
Stan H.
Answer by robertb(5830)  (Show Source):
You can put this solution on YOUR website! Use the vertex form :  .
Then  . Then plug in the respective coordinates of the given point (-5,6) into this equation to solve for a. Think you can manage from there already.
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