How do you graph the parabola #y=(x+4)^2+2# using vertex, intercepts and additional points?

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Answer 1 · 2018-07-28T18:29:25

Given equation of parabola:

#y=(x+4)^2+2#

#(x+4)^2=y-2#

Comparing above equation with the standard form of vertical parabola: #(x-x_1)^2=4a(y-y_1)#, we get

#x_1=-4, y_1=2, a=1/4#

The vertex of above parabola is at #(x_1, y_1)\equiv(-4, 2)#

Axis of symmetry: #x-x_1=0#

#x+4=0#

Since, the parabola is upward & above x-axis hence it wouldn't intersect the x-axis.

Now, to find y-intercept we set #x=0# in given equation, as follows

#y=(0+4)^2+2#

#y=18#

hence the parabola intersects the y-axis at #(0, 18)#

Steps to draw the graph of parabola:

1) Draw the axis of symmetry #x+4=0# or #x=-4#

2) Specify the vertex #(-4, 2)# on the axis of symmetry

3) Draw a free hand symmetric graph of upward parabola symmetric about #x=-4# such that its one arm intersects the y-axis at #(0, 18)#.

Thanks!