What is the vertex form of #y=2x^2+4x-5#?

1 Answer
Jul 26, 2016

#y=color(green)(2)(x-color(red)(""(-1)))^2+color(blue)(""(-8))#

Explanation:

Given:
#color(white)("XXX")y=2x^2+4x-5#

Remember that the vertex form 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))#

Extracting the the #color(green)(m)# factor from the given equation
#color(white)("XXX")y=color(green)(2)(x^2+2x)-5#

Complete the square
#color(white)("XXX")y=color(green)(2)(x^2+2xcolor(purple)(+1))-5-color(green)(2)*color(purple)(1))#

Rewrite with a squared binomial and simplified constant
#color(white)("XXX")y=color(green)(2)(x-color(red)(""(-1)))^2+color(blue)(""(-8))#
which is the vertex form with vertex at #(color(red)(-1),color(blue)(-8))#

Here is a graph of the original equation for verification purposes:
graph{2x^2+4x-5 [-10.83, 11.67, -10.08, 1.17]}