What is the the vertex of #y = (x -3)^2-9x+5 #?

1 Answer
Oct 5, 2017

Vertex at : #(7 1/2,-42 1/4)#

Explanation:

Given
#color(white)("XXX")y=(x-3)^2-9x+5#

Expanding:
#color(white)("XXX")y=x^2-6x+9-9x+5#

#color(white)("XXX")y=x^2-15x+14#

We can proceed from here in 2 ways:

  • by converting this into vertex form through "completing the square" method
  • using the axis of symmetry (below)

Using the axis of symmetry
Factoring we have
#color(white)("XXX")y=(x-1)(x-14)#
which implies #y=0# (the X-axis) when #x=1# and when #x=14#

The axis of symmetry passes through the midpoint between the zeros
i.e. the axis of symmetry is #x=(1+14)/2= 15/2#

Note that the axis of symmetry also passes through the vertex;
so we can solve the original equation (or more easily our factored version) for the value of #y# where the equation and the axis of symmetry intersect:
#color(white)("XXX")y=(x-1)(x-14)# for #x=15/2#

#color(white)("XXX")rarr y=(15/2-1)(15/2-14)=13/2 * (-13/2))=-169/4#

So the vertex is at #(15/2,-169/4)=(7 1/2,-42 1/4)#

We can verify this result with a graph of the original equation:

graph{(x-3)^2-9x+5 [-0.016, 14.034, -45.34, -38.32]}