Asked by Anonymous
                A quadratic function has its vertex at the point (-9,8). The function passes through the point (9,4). When written in vertex form, the function is f(x)= a(x-h)^2 +k, where
a)
h)-9
k)8
4=a(9)^2+8
4=a(9+9)^2+8
4=a(18)^2+8
4=a(332)
a=0.01204
I was able to find both h & k. but the answer I got for a is incorrect, please help me
            
        a)
h)-9
k)8
4=a(9)^2+8
4=a(9+9)^2+8
4=a(18)^2+8
4=a(332)
a=0.01204
I was able to find both h & k. but the answer I got for a is incorrect, please help me
Answers
                    Answered by
            Reiny
            
    from vertex,
y = a(x+9)^2 + 8
but (9,4) lies on it, so
4 = a(18^2) + 8
324a = -4
a = -4/324 = -1/81
etc
you messed up here:
4=a(18)^2+8
4=a(332) , you added 8 to 324a , you can't add unlike terms
    
y = a(x+9)^2 + 8
but (9,4) lies on it, so
4 = a(18^2) + 8
324a = -4
a = -4/324 = -1/81
etc
you messed up here:
4=a(18)^2+8
4=a(332) , you added 8 to 324a , you can't add unlike terms
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