Question

How do you find the vertex and intercepts for `y=(x+3)(x+5)`?

Answer 1

y-intercept: `15`
x-intercepts: `(-3)` and `(-5)`
vertex: `(-4,-1)`

Explanation:

Starting with the intercepts:

The y-intercept is the value of `y` when `x=0`.
`y=(0+3)(0+5)=15`

The x-intercept(s) is/are the values of `x` when `y=0`
`0=(x+3)(x+5)`
`rArr x=-3` or `x=-5`

Determining the vertex.
Method 1:

For a parabola with a vertical axis,
the x coordinate of the vertex will be half way between the two x-intercepts.
i.e. the x-coordinate of the vertex will be `((-3)+(-5))/2 = -4`
By substituting `(-4)` for `x` in the given equation
`color(white)("XXX")y=(-4+3)(-4+5)=-1`
So the vertex will be at `(-4,-1)`

Method 2:

Convert the given equation into vertex form: `y=m(x-color(red)(a))^2+color(green)(b)`
for a parabola with vertex at `(color(red)(a),color(green)(b))`
`color(white)("XXX")y=(x+3)(x+5)`
`color(white)("XXX")rarr y=x^2+8x+15`
`color(white)("XXX")rarr y=x^2+8xcolor(blue)(+16) + 15 color(blue)(-16)`
`color(white)("XXX")rarr y=(x+4)^2+(-1)`
`color(white)("XXX")rarr y = (x- (color(red)(-4)))^2+(color(green)(-1))`