Question

How do you graph and label the vertex and axis of symmetry of `y=(x+4)(x+2)`?

Answer 1

Vertex is `(-3,-1)` and axis of symmetry is `x+3=0`

Explanation:

`y=(x+4)(x+2)` is the equation of parabola in intercept form and `-2` and `-4` are the `x` intercepts of parabola.

To find vertex and axis of symmetry weshould convert it into vertex form `y=a(x-h)^2+k`, where `(h,k)` is vertex and axis of symmetry is `x+h=0`. Hence let us convert `y=(x+4)(x+2)` into vertex form as indicated below.

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

= `x^2+6x+8`

= `x^2+6x+9-1`

= `(x+3)^2-1`

Hence vertex is `(-3,-1)` and axis of symmetry is `x+3=0`

graph{(x+3)(x^2+6x+8-y)((x+3)^2+(y+1)^2-0.02)=0 [-12.62, 7.38, -4.84, 5.16]}