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

1 Answer
Alan P.
Apr 6, 2016

#y=2(x-(-5/4))^2+(-6 1/8)#

Explanation:

General vertex form:
#color(white)("XXX")y=m(x-color(blue)(a))^2+color(orange)(b)# with vertex at #(color(blue)(a),color(orange)(b))#

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

Extract "spread factor" #m#
#color(white)("XXX")y=2(x^2+5/2x)-3#

Complete the square
#color(white)("XXX")y=color(red)(2)(x^2+5/2xcolor(red)(+(5/4)^2))-3-color(red)(2(5/4)^2)#

Write as a squared binomial and simplify the constant
#color(white)("XXX")y=2(x+5/4)^2-6 1/8#

Re-write to match signs of standard general form:
#color(white)("XXX")y=2(x-color(blue)(color(white)("")(-5/4)))^2+color(orange)(color(white)("")(-6 1/8))#

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