Question 922443: Find the equation of a parabola with roots -3 and 6 and contains the point (1,10). I have tried using k- a( x-h)^2 but as I have no vertex???? I do think 1.5 is the x value of vertex.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Find the equation of a parabola with roots -3 and 6 and contains the point (1,10). I have tried using k- a( x-h)^2 but as I have no vertex???? I do think 1.5 is the x value of vertex.
:
Using the ax^2 + bx + c = y, find a,b,c using the given points
x=-3; y=0
9a - 3b + c = 0
x=6; y=0
36a + 6b + c = 0
x=1; y=10
a + b + c = 10
:
Subtract the 1st equation from the 2nd equation
36a + 6b + c = 0
9a - 3b + c = 0
-------------------subtraction eliminates c
27a + 9b = 0
Likewise, subtract 3rd equation from the 2nd equation
36a + 6b + c = 0
a + b + c = 10
------------------eliminates c again
35a + 5b = -10
simplify, divide by 5
7a + b = -2
multiply by 9, subtract the other 2 unknown equation
63a + 9b = -18
27a + 9b = 0
--------------subtraction eliminates b, find a
36a = -18
a = -18/36
a = -.5
Find b using the equation 7a + b = -2
7(-.5) + b = -2
-3.5 + b = -2
b = -2 + 3.5
b = 1.5
Find c using the 3rd equation a + b + c = 10
-.5 + 1.5 + c = 10
c = 10 - 1
c = 9
Our equation: y = -.5x^2 + 1.5x + 9
:
Check using the given point 1,10
y = -.5(1^2) + 1.5(1) + 9
y = -.5 + 1.5 + 9
y = 10
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