What are he intercepts of #y=(x+1)^2-2#?

1 Answer
Jul 25, 2018

Answer:

The #x#-intercepts are at #(sqrt2-1)# and #(-sqrt2-1)# and the #y#-intercept is at #(0, -1)#.

Explanation:

To find the #x#-intercept(s), plug in #0# for #y# and solve for #x#.

#0 = (x+1)^2 - 2#

Add #color(blue)2# to both sides:
#2 = (x+1)^2#

Square root both sides:
#+-sqrt2 = x+1#

Subtract #color(blue)1# from both sides:
#+-sqrt2 - 1 = x#

Therefore, the #x#-intercepts are at #(sqrt2-1)# and #(-sqrt2-1)#.

To find the #y-#intercept, plug in #0# for #x# and solve for #y#:
#y = (0+1)^2 - 2#

Simplify:
#y = 1^2 - 2#

#y = 1 - 2#

#y = -1#

Therefore, the #y#-intercept is at #(0, -1)#.

Hope this helps!