SOLUTION: The dimensions of the facade of the house, should be 7m wide and 4 ms high, it should be a parabola, use the origin of the parabola as one corner. Find the quadratic equation to th

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: The dimensions of the facade of the house, should be 7m wide and 4 ms high, it should be a parabola, use the origin of the parabola as one corner. Find the quadratic equation to th      Log On


   



Question 956636: The dimensions of the facade of the house, should be 7m wide and 4 ms high, it should be a parabola, use the origin of the parabola as one corner. Find the quadratic equation to that.
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3 points are (0,0), (3.5,4) and (7,0)
y = ax^2 + bx + c
At x = 0, y = 0 --> c = 0
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At x = 7, y = 0
49a + 7b = 0
7a + b = 0
b = -7a
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At x = 3.5, y = 4
12.25a + 3.5b = 4
Sub for b
12.25a - 24.5a = 4
-12.25a = 4
a = -16/49
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b = 16/7
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And then they got
y=16x^2/49+16x/7
I do not understand the process here, I am practicing for my exam and this on the practice exam, could someone please explain the process clearly to me? That would be greatly appreciated

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
(0,0)
0=0%2B0%2Bc
highlight%28c=0%29
.
.
.
(7/2,4)
4=a%287%2F2%29%5E2%2Bb%287%2F2%29
4=%2849%2F4%29a%2B%287%2F2%29b
1.49a%2B14b=16
.
.
.
(7,0)
0=a%287%29%5E2%2Bb%287%29
2.49a%2B7b=0
Subtract eq. 2 from eq. 1,
49a%2B14b-49a-7b=16-0
7b=16
highlight%28b=16%2F7%29
Then,
49a%2B7%2816%2F7%29=0
49a%2B16=0
49a=-16
highlight%28a=-%2816%2F49%29%29
So then,
y=-%2816%2F49%29x%5E2%2B%2816%2F7%29x