How do you find a quadratic function whose vertex is at the point (2,9) and has the given x intercepts (0.5,0) & (3.5,0)?

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Answer 1 · 2015-06-24T18:44:49

Express in vertex form as #f(x) = a(x-2)^2+9#
and solve for #a# to find #a = -4#, so

#f(x) = -4(x-2)^2+9 = -4x^2+16x-7#

In vertex form, the function takes the form:

#f(x) = a(x-2)^2+9# for some constant #a#

#0 = f(0.5) = a(0.5-2)^2+9 = 2.25a+9#

So #2.25a = -9# and #a = -9/2.25 = -4#

So

#f(x) = -4(x-2)^2+9#

#=-4(x^2-4x+4)+9#

#=-4x^2+16x-16+9#

#=-4x^2+16x-7#