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The graph of the function f(x)=ax^2+bx+c has its vertex at (0,1) and passes through the point (1,9) find a b and c

 Mar 15, 2020
 #1
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The quadratic is symmetrical about its vertex, so, if it passes through (1, 9) it also passes through (-1,9), since the x-value of the vertex is at 0.

 

So you have three equations:

 

a*(-1)2 +b*(-1) + c = 9

a*12 + b*1 + c = 9

a*02 + b*0 + c = 1

 

You should be able to obtain a, b and c from these three equations.

 Mar 15, 2020

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