How do you find the vertex of #y= (x-3)(x+3)#?

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Answer 1 · 2015-07-15T12:54:21

Construct the vertex using part geometry and part algebra to find #(0, -9)#...

This is a vertical parabola that cuts the #x#-axis at #(-3, 0)# and #(3, 0)#.

Since parabolas are symmetrical, its axis must be midway between these, namely the vertical line #x=0#.

The axis intersects the parabola at the vertex, so substitute #x=0# into the equation to get:

#y = (0-3)(0+3) = -3 * 3 = -9#

that is the point #(0, -9)#.

graph{(x-3)(x+3) [-19.91, 20.09, -12.08, 7.92]}

Answer 2 · 2015-07-15T13:01:11

Multiply out to find #y = (x-3)(x+3) = x^2-9#

#x^2 >= 0# for all #x#, with minimum when #x=0#,

giving #y = x^2-9 = 0^2-9 = -9#, that is #(0, -9)#

#(x-3)(x+3)# is recognisable as a factorisation of #x^2-3^2 = x^2-9#

(use the difference of squares identity: #a^2-b^2 = (a-b)(a+b)#)

So #y = (x-3)(x+3) = x^2-9#

This takes minimum value when #x^2 = 0#, that is when #x=0#

When #x = 0#, we have #y = x^2-9 = 0-9 = -9#

So the vertex is at #(0, -9)#