How do you find a quadratic function whose vertex is at the point (2,9) and has the given x intercepts (-1,0) & (5,0)?

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Answer 1 · 2015-06-02T21:08:52

From the #x# intercepts, we can deduce:

#f(x) = a(x+1)(x-5)# for some constant #a#

Then from the vertex point we get:

#9 = f(2) = a(2+1)(2-5) = -9a#

So #a=-1#

#f(x) = -(x+1)(x-5) = -(x^2-4x-5)#

#= -x^2+4x+5#