Question 96842: Y=p-(x+q)^ crosses the y-axis at the point (0,1) and x axis at (4,0) find the value of p and q [PS: ^=2]
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Y=p-(x+q)^ crosses the y-axis at the point (0,1) and x axis at (4,0) find the value of p and q [PS: ^=2]
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Y=p-(x+q)^2
Substitute (0,1) to get:
1 = p-q^2
Substitute (4,0) to get:
0=p-(4+q)^2
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Rearrange equations:
p = 1+q^2
p = (4+q)^2
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EQUATION:
1+q^2 = (4+q)^2
1+q^2 = 16+8q+q^2
8q=-15
q = -15/8
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Substitute that into p = 1+q^2 to solve for "P":
p = 1 + (-15/8)^2
p = (64 + 225)/64
p = 289/64
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Cheers,
Stan H.
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