Question

How do you find the vertex and intercepts for `y=x^2+4x-5`?

Answer 1

Vertex: `(-2,-9)`
y-intercept: `(-5)`
x-intercepts: `(-5) and (+1)`

Explanation:

To begin, let's convert the given equation
`color(white)("XXX")y=x^2+4x-5`
into "vertex form": `y=m(x-a)^2+b` (with vertex at `(a,b)`)

Completing the square:
`color(white)("XXX")y=x^2+4xcolor(green)(+4)-5color(green)(-4)`

Simplifying and writing as a squared binomial:
`color(white)("XXX")y=(x+2)^2-9` (** this is the form I will use for the intercepts later)
or
`color(white)("XXX")y=1(x-(-2))^2+(-9)`
which is in vertex form with vertex at `(-2,-9)`

The y-intercept is the value of `y` when `x=0`
`color(white)("XXX")y=(0+2)^2-9=-5`

The x-intercepts are the values of `x` when `y=0`
`color(white)("XXX")0=(x+2)^2-9`

`color(white)("XXX")rarr (x+2)^2=9`

`color(white)("XXX")rarr x+2 = +-3`

`color(white)("XXX")rarr x= -5 or +1`

graph{x^2+4x-5 [-12.37, 10.13, -10.305, 0.945]}