Question

How do you find the vertex and the intercepts for `y=-4x^2+24x`?

Answer 1

Y-intercept is `y=0`.
X-intercepts at `x=0` and `x=6`
Vertex at `(3, 36)`

Explanation:

Y-intercept.

This is the place the graph cuts the y-axis, which is the line `x=0`.

So `"let " x=0`

`y=-4*0^2-24*0`
`y=0`
`:. "y-intercept at " (0,0)`
The y-intercept is 0.

OK, you don't need to sub in x=0 physically to work this out, but it gets you into the habit. In a quadratic of the from `y=ax^2+bx+c`, the y-intercept is at `c`.

X-intercept

This is where the graph cuts the line `y=0`

So `"let "y=0`
`0=4x^2-24x`
`0=4x(x-6)`
`:. x=0 " or " x=6`
So the x-intercepts are at `x=0` and `x=6`

Vertex

For this, we need to complete the square.

`y=-4x^2+24`
`=-4(x^2-6)`
`=-4([x-3]^2-9)`
`=-4[x-3]^2+36`

So the vertex is at `(3, 36)`

Alternatively, if you know any calculus, you could differentiate this expression and set the derivative to 0. If you don't, don't worry - completing the square works fine, although sometimes fractions can be awkward.

Here is the graph `y=-4x^2+24x`

graph{-4x^2+24x [-52.6, 51.4, -4.9, 47.1]}