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

1 Answer
Apr 4, 2018

Answer:

y-intercept
#y=-5#

x-intercepts
#x_1 =-5#
#x_2 =1#

Explanation:

an x-intercept is a point where #y=0#, and
a y-intercept is a point where #x=0#

x-intercepts: #0=x^2 + 4x -5#

#x^2 + 4x -5+ (4/2)^2=0+ (4/2)^2# complete the square
#(x + 2)^2 -5=4#
#(x + 2)^2 =9#
#x + 2=± 3#
#x =-2± 3#

#x_1 =-5#
#x_2 =1#

y-intercept: #y=(0)^2 + 4(0) -5=-5#

Check the graph

graph{x^2 + 4x - 5 [-9.96, 10.04, -8.72, 1.28]}