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

1 Answer

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!