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.
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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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{{{graph(400,300,-10,10,-10,10,(-1/4)x^2+3)}}}}
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Cheers,
Stan H.