How do you find a quadratic function whose vertex is at the point (2,9) and has the given x intercepts (-2,0) & (7,0)?

1 Answer
Jun 2, 2015

Given the intercepts, the quadratic must be of the form:

#f(x) = a(x+2)(x-7)# for some constant #a#

Now we know that #f(2) = 9#, so:

#9 = f(2) = a(2+2)(2-7) = -20a#

Divide both sides by #-20# to get:

#a = -9/20#

So

#f(x) = -9/20(x+2)(x-7) = -9/20(x^2-5x-14)#

But there is a problem: #(2, 9)# is not the vertex of this quadratic. The vertex will have #x = 5/2 = (-2+7)/2# where

#f(5/2) = -9/20(5/2+2)(5/2-7)#

#=-9/20(5/2+4/2)(5/2-14/2)#

#=-9/20*9/2*-9/2#

#=729/80 = 9.1125#

So the problem as stated seems to be incorrect in one of the following ways:
(1) One of the intercepts was given incorrectly. For example, #(-2, 0)# instead of #(-3, 0)#; or #(-7, 0)# instead of #(-6, 0)#.
(2) The point #(9, 2)# is just a point through which the quadratic passes - not necessarily a vertex.
(3) The parabola is tilted from the vertical, described by a 2nd order equation in #x# and #y#. That would not express #y# as a quadratic function of #x# or vice-versa.