Y=x^{2} +2x-8 determine the axis of symmetry, the vertex point and find the roots by factoring

We have axis of symmetry in

The vertex point is

The roots by factoring are

Step-by-step explanation:

We have the function:

Factoring the quadratic equation:

When

The vertex point is

From the equation, we have

We also got the axis of symmetry

In order to find just use the :

Once , we have the minimum at

x = -1  axis of symmetry

(-1,-9) vertex

x = -4           x=2   are the roots

Step-by-step explanation:

y = x^2 + 2x - 8

The axis of symmetry is

h = -b/2a

h = -2/ (2*1) = -2/2 =-1

x = -1

The x coordinate of the vertex is at the axis of symmetry

To find the y value substitute into the function

y = (-1)^2 +2(-1) -8

y = 1-2-8

y = -9

The vertex is at ( -1,-9)

Factor

What 2 number multiply to -8 and add to 2

4*-2 = -8

4+-2 = 2

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

Using the zero product property to find the roots

0 = ( x+4) (x-2)

x+4 =0    x-2 =0

x = -4           x=2   are the roots

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