How do you graph #y=(x+1)^2 - 4#?

1 Answer
Jul 13, 2018

vertex: #(-1, -4); " axis of symmetry: "x = -1#

#y#-intercept: #(0, -3); " "x"-intercepts: "(-3, 0), (1, 0)#

Explanation:

Given: #y = (x + 1 )^2 - 4#

This is an equation of a parabola in vertex form:

#y = a(x - h)^2 + k; " where the vertex is " (h, k)# and

the axis of symmetry is #x = h# and #a# is a constant.

Find the vertex and axis of symmetry:

vertex: #(-1, -4); " axis of symmetry: "x = -1#

Find the y-intercept by setting #x = 0#:

#y = (0+1)^2 - 4 = 1-4 = -3 => (0, -3)#

Find x-intercepts by setting #y# = 0 and factoring or using the quadratic formula or completing the square:

#0 = (x + 1 )^2 - 4 = (x^2 + 2x + 1) - 4#

#x^2 + 2x -3 = (x - 1)(x + 3) = 0#

#x#-intercepts: #(-3, 0), (1, 0)#

To graph, plot the intercepts and the vertex. You can use the technique of point-plotting to add additional points.

Since #x# is an independent variable, you can select different #x# values and calculate the corresponding #y# values.

#ul(" "x" "|-4" "|-2" "|" "2" "|)#
#" "y" "|" "5" "|-3" "|" "5" "|#

graph{(x + 1 )^2 - 4 [-10, 10, -5, 5]}