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

Answer 1

Ethan Adkins

Be able to recognize the vertex form of a parabola.

A parabola's vertex form is:

#y=(x-h)^2+k#

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

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

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

The vertex's x-coordinate is where the axis of symmetry passes through:

#x = -2#

graph{(x+2)^2-8 [–20,–20,–10]}

Hope that helps

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Answer 2

Kennedy Adams

To find the vertex of the parabola represented by the function ( f(x) = (x+2)^2 - 8 ), we need to identify the values of ( h ) and ( k ) in the vertex form ( f(x) = a(x - h)^2 + k ). In this case, ( h = -2 ) and ( k = -8 ). Therefore, the vertex is at ( (-2, -8) ). The axis of symmetry is the vertical line passing through the vertex, which is ( x = -2 ). To graph the parabola, plot the vertex at ( (-2, -8) ), and then use the symmetry to find additional points on the graph. Since the coefficient of ( x^2 ) is positive, the parabola opens upwards.

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Answer from HIX Tutor

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