Question 135108: Ok my problem is a graph and im supposed to find the quadractic function and put it in standard form. The graph has two points on it which are vertex(0,3) and another point of (2,4). so today we went over in class how to solve problems like these so i put the information i had into f(x)= a(x-h)∧2 + k. I get all the way down to solving it but when i graph it. It doesnt graph out right which makes me think im doing it wrong. I know a is supposed to be negative but when i solved it, it came out positive so that was also wrong. I just dont know what to do.
Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website! Using  means the vertex is at (h,k). Since you are given the vertex of (0,3), that means the equation must be  , or more simply:  . For the equation to pass through the point (2,4),  , so  =>  =>  , and the equation must be:
You can see that "a" in the general equation cannot be negative in this case because that would mean the parabola would open downward. A downward opening parabola with a vertex at (0,3) cannot possibly contain the point (2,4).
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! Ok my problem is a graph and im supposed to find the quadractic function and put it in standard form. The graph has two points on it which are vertex(0,3) and another point of (2,4). so today we went over in class how to solve problems like these so i put the information i had into f(x)= a(x-h)∧2 + k. I get all the way down to solving it but when i graph it. It doesnt graph out right which makes me think im doing it wrong. I know a is supposed to be negative but when i solved it, it came out positive so that was also wrong. I just dont know what to do.
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f(x)= a(x-h)^2 + k.
You are given x,y,h, and k; so solve for "a"
If you know the parabola opens down a must be
negative.
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4 = a(2-0)^2+3
4 = 4a + 3
4a = 1
a = 1/4, but you know it is -1/4
EQUATION:
y = (-1/4)(x-0)^2 + 3
y = (-1/4)x^2+3
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Cheers,
Stan H.
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