Question

How do you find x-intercepts, coordinates of vertex for parabola `y = x^2 - 4x - 12`?

Answer 1

`y = x^2-4x-12`

Part 1: x-intercepts
The x-intercepts occur at the points on the function where `y=0`
So, we need to solve
`x^2-4x-12 = 0`
The left side factors fairly easily into:
`(x-6)(x+2)=0`
So solution occur when
`x-6 = 0 rarr x=6`
and
`x+2= 0 rarr x=(-2)`
So the x-intercepts are at `(0,6)` and `(0,-2)`

Part 2: vertex of the parabola
The vertex of a simple quadratic parabola occurs when the derivative of the quadratic is equal to `0`.
The derivative of the given quadratic is
`(dy)/(dx) = 2x -4`
By observation, this is equal to `0` when `x=2`
When `x=2` the original equation becomes
`y = (2)^2 -4(2)-12`
`y = -16`
Therefore the vertex of this parabola is at `(2,-16)`