Question

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

Answer 1

See below

Explanation:

For the vertex, we need to get our equation in standard form. Using FOIL, it can be rewritten as `y=x^2-2x-8`. To find its x-coordinate, use the formula `x=-b/(2a)`:

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

The y-coordinate is found by plugging our answer back into the equation:

`y=(1)^2-2(1)-8`
`y=1-2-8=-7`

Vertex: `(1,-7)`

By "intercepts", I believe you mean both the x-intercepts and the y-intercept. For the x-intercepts, we can use the factored equation, set `y` to `0`, and solve:

`0=(x-4)(x+2)`
`x-4=0`
`x=4`
`x+2=0`
`x=-2`

X-intercepts: `(4,0), (-2,0)`

To find the y-intercept, we will use the standard form of our equation, set `x` to `0`, and solve:

`y=0^2-2(0)-8`
`y=-8`

Y-intercept: `(0,-8)`

Hope this helped!