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A vertical parabola has vertex (4,1) and Y intercept 5. What are its X intercepts?
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y= A(x-4)^2 + 1 is the general form of such a parabola
we know that when x=0, y= 5
so we can find A
5 = 16A +1 A= (1/4)
so y= (1/4) (x-4)^2 + 1 x intercepts, set y=0
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(x-4)^2 = -4 , this has no solution! , no x intercepts
Thank you
how did you get -4?
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