Question

How do you find the axis of symmetry, vertex and x intercepts for `y=x^2+x-6`?

Answer 1

the vertex will be `(-1/2,-25/4)`
the axis of symmetry is simply `X=-1/2`

`x" "`intercepts `x=-3,2`

Explanation:

The first thing to do is to complete the square

`y=x^2+x-6`

add half the coefficient of `x` and square it, and subtract to balance

`y=color(red)(x^2+x+(1/2)^2)-6-(1/2)^2`

the terms in red give a perfect square

`y=color(red)((x+1/2)^2)-25/4`

the vertex will be `(-1/2,-25/4)`

basically this is the translation of `y=x^2` to this graph.

graph{x^2+x-6 [-10, 10, -5, 5]}

the axis of symmetry is simply `X=-1/2`

the `x` intercepts we solve

`(x+1/2)^2-25/4=0`

`=>(x+1/2)^2=25/4`

`x+1/2=+-sqrt(25/4)`

`x+1/2=+-sqrt(25/4)`

`x+1/2=+-5/2`

`:.x=+-5/2-1/2`

`x=-3,2`

Answer 2

Axis of symmetry: `x = -1/2`
Vertex is at `(-1/2, -6 1/4)`

`x`-intercepts are at `-3 and 2`

Explanation:

The quadratic is in the form `y = ax^2 +bx +c`

`y = x^2 +x -6" "a = 1, b= 1 and c = -6`

The axis of symmetry is found from: `x = (-b)/(2a)`

`x = (-1)/(2(1)) =-1/2`

The vertex lies on the axis of symmetry.
You have the `x`-value, substitute to find the `y`-value:

`y = (-1/2)^2 -1/2-6" "= -6 1/4`

Vertex is at `" "(-1/2, -6 1/4)`

To find the `x`-intercepts, set `y=0` and solve

`x^2 +x-6 =0" "larr` factorise

`(x+3)(x-2)=0`

`x = -3 and x = +2`

(The average of the x-intercepts confirms the axis of symmetry)

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