How do you find the vertex and axis of symmetry, and then graph the parabola given by: $f (x) = {(x + 2)}^{2} - 8$ $f(x)=(x+2)^2-8$ ?

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Answer 1 · 2015-09-27T19:37:52

Be able to recognize the vertex form of a parabola.

The vertex form for a parabola is:

$y = {(x - h)}^{2} + k$ $y=(x-h)^2+k$

Where, $(h, k)$ $(h,k)$ is the vertex.

In this problem, $h = - 2$ $h = -2$ and $k = - 8$ $k=-8$

So, the vertex $= (- 2, - 8)$ $=(-2,-8)$

The axis of symmetry passes through the x-coordinate of the vertex:

$x = - 2$ $x = -2$

graph{(x+2)^2-8 [-20, 20, -10, 10]}

Hope that helps

How do you find the vertex and axis of symmetry, and then graph the parabola given by: f(x)=(x+2)2−8 f(x)=(x+2)^2-8?