Question

How do you find the axis of symmetry, vertex and x intercepts for `y=x^2+2x`?

Answer 1

Axis of Symmetry : `x = -1`

Vertex = ( - 1, - 1 )

X-Intercept = There are two values : ( - 2 ) and Zero

Explanation:

We are given the Quadratic `y = x^2 + 2 * x`

We will now find the Vertex Form.

In general, Vertex Form is written as

`y = f( x ) = a*(x - h) ^ 2 + k`

Our Vertex will be ( h, k )

`f( x ) = [ x ^ 2+ 2 * x + ( ) ] - ( )`

(Divide the coefficient of the x-term by 2 and square it: (2/2) ^ 2, to add and subtract)

`f( x ) = [ x ^ 2+ 2 * x + ( 1 ) ] - ( 1 )`

We can write the above as

`f( x ) = ( x + 1) ^ 2 - 1`

Now, `h = (-1)` and `k = (- 1)`, in the general Vertex Form.

We now have the Vertex Form of the Quadratic

In our problem situation, our Vertex is ( - 1, - 1 )

Axis of Symmetry is given by: x = h

Hence, x = ( - 1 ) is the required Axis of Symmetry

To find the x-Intercept, equate f( x ) = 0

`0 = (x + 1) ^ 2 - 1`

`(x + 1) ^ 2 - 1 = 0`

`(x + 1) ^ 2 = 1`

`(x + 1) = +- sqrt(1)`

This gives us ( - 2) and Zero as the two x-Intercepts.