Question

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

Answer 1

We have axis of symmetry in
`x=-1`

The vertex point is
`(-1, -9)`

The roots by factoring are

`x=-4`

`x=2`

Explanation:

We have the function:

`f(x)=x^2+2x-8`

Factoring the quadratic equation:

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

When
`y=0`

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

`x=-4`

`x=2`

The vertex point is
`(h, k)`

From the equation, we have
`a=1,\:b=2 \text{ and }\:c=-8`

`$h=-(b)/(2a)$`

`$h=-(2)/(2\cdot \:1)$`

`h=-1`

We also got the axis of symmetry

In order to find
`k` just use the
`$h=x_(vetex)$`:

`k=(-1)^2+2(-1)-8\\k=1-2-8\\k=-9`

Once
`a>0`, we have the minimum at
`(-1, -9)`

Answer 2

x = -1 axis of symmetry

(-1,-9) vertex

x = -4 x=2 are the roots

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