Question

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

Answer 1

Axis of sym at `x= -1`, vertex at (-1,1)

`x-`intercepts at -2 and 0

`y`-intercept at (0,0)

Explanation:

Let's find the axis of symmetry first, it is an easy way to find the vertex.
`color(red)("axis of symmetry:" x= (-b)/(2a)) = (-(-2))/(2(-1)) = -1`
`x=-1`

The vertex lies on the axis of symmetry:

If `x=-1, " find " y = -(-1)^2 -2(-1) = -1+2 = 1`

`color(blue)("The vertex is at " (-1,1))`

Intercepts:
For y-int: `x=0 rarr y=0`

For x-intercepts : `y = 0`

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

`0 = -x(x+2) rArr x=0, or x=-2`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the completing the square method to find the vertex:

`y = -x^2 -2x`
`y = -1(x^2 +2x)`
`y=-1[color(magenta)(x^2 +2x +1)-1]`
`y =-1[color(magenta)((x+1)^2) -1]`
`y = -(xcolor(blue)(+1))^2 color(blue)(+1)`

`color(blue)("The vertex is at " (-1,1))`

Confirm these points on the graph below.

graph{-x^2-2x [-3.5, 1.5, -1.4, 1.1]}