How do you find the vertex and intercepts for #y=x^2+4x-5#?

1 Answer
Feb 25, 2016

Vertex: #(-2,-9)#
y-intercept: #(-5)#
x-intercepts: #(-5) and (+1)#

Explanation:

To begin, let's convert the given equation
#color(white)("XXX")y=x^2+4x-5#
into "vertex form": #y=m(x-a)^2+b# (with vertex at #(a,b)#)

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

Simplifying and writing as a squared binomial:
#color(white)("XXX")y=(x+2)^2-9# (** this is the form I will use for the intercepts later)
or
#color(white)("XXX")y=1(x-(-2))^2+(-9)#
which is in vertex form with vertex at #(-2,-9)#

The y-intercept is the value of #y# when #x=0#
#color(white)("XXX")y=(0+2)^2-9=-5#

The x-intercepts are the values of #x# when #y=0#
#color(white)("XXX")0=(x+2)^2-9#

#color(white)("XXX")rarr (x+2)^2=9#

#color(white)("XXX")rarr x+2 = +-3#

#color(white)("XXX")rarr x= -5 or +1#

graph{x^2+4x-5 [-12.37, 10.13, -10.305, 0.945]}