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]}
