How do you write the vertex form equation of the parabola #(x+5)(x+4)#?

1 Answer
Apr 24, 2016

#y=(x-(-9/2))^2+(-1/4)#

Explanation:

The general vertex form for a parabola (with a vertical axis of symmetry) is
#color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)#
with vertex at #(color(red)(a),color(blue)(b))# (#color(green)(m)# can be thought of as a parameter that effects the "spread" of the parabola).

Given:
#color(white)("XXX")y=(x+5)(x+4)#

This can be rewritten as:
#color(white)("XXX")y=x^2+9x+20#

Completing the square:
#color(white)("XXX")y=x^2+9xcolor(brown)(+(9/2)^2)+20color(brown)(-(9/2)^2)#

#color(white)("XXX")y=(x+9/2)^2+(80/4-81/4)#

#color(white)("XXX")y=color(green)(1)(x-color(red)(""(-9/2)))^2+color(blue)(""(-1/4))#
with vertex at #(color(red)(-9/2),color(blue)(-1/4))#

(Note that when #color(green)(m)=1# it is often omitted.)

graph{(x+5)(x+4) [-5.95, -0.474, -1.327, 1.41]}