Question

How do you graph and label the vertex and axis of symmetry `y=x^2+2x`?

Answer 1

See below.

Explanation:

To graph we need to find significant points.

The y axis intercept occurs where `x=0`:

So: `y= 0^2+2(0)=> y=0`

`x` axis intercepts occurs where `y = 0`

Factor `x^2+2x`

`(x^2+2x)=>x(x+2)`

`x(x+2)=0=> x=0, x=-2`

So parabola intercepts `x` axis at the points:

`(0,0)` and `(-2 ,0)`

Since you need to label vertex and axis of symmetry you will need to arrange `y=x^2+2x` into the form:

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

Where `h` is the axis of symmetry and `k` is the maximum/minimum value.
To do this place a bracket around the terms containing the variable.

`(x^2+2x)`

Factor out the coefficient of `x^2` if this is not 1:

Add the square of half the coefficient of `x` inside the bracket, and subtract it outside the bracket:

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

Make `(x^2 +2x +(1)^2)` into the square of a binomial:

`(x+1)^2`

We now have:

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

We can now see that the axis of symmetry is `-1`

Minimum value is `-1`

We know this is a minimum value because the coefficient of `x^2` was positive, so the parabola will be this way up `uuu`

Since the coordinates of the vertex is ( axis of symmetry, max/min value).

We have:

`(-1 ,-1)`

Graph of `y=x^2+2x`

graph{y=x^2+2x [-5, 5, -3, 5.625]}