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

1 Answer
Sep 29, 2017

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]}