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

1 Answer
Feb 1, 2016

vertex at #(-2,-9)#

Explanation:

Often the simplest way to do this is to convert the given equation into "vertex form":
#color(white)("XXX")y=(x-a)^2+b# with its vertex at #(a,b)#

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

Completing the square:
#color(white)("XXX")y=x^2+4xcolor(blue)(+4)-5color(blue)(-4)#

Re-writing as a squared binomial and simplified constant
#color(white)("XXX")y=(x+2)^2-9#

Modifying signs into explicit vertex form:
#color(white)("XXX")y=(x-(-2))^2+(-9)#

If you have access to some graphing software, it can help verify that the answer is reasonable by graphing the original equation.
graph{x^2+4x-5 [-8.91, 11.09, -9.59, 0.41]}