Question

What is the vertex of `y=4x-x^2`?

Answer 1

The vertex is at the point `(2, 4)`.

Explanation:

If you write your quadratic in standard form, (`ax^2+bx+c`) then the `x`-coordinate of the vertex is calculated by `(-b)/(2a)`.

Let's convert that equation to standard form:

`y=4x-x^2`

`quad=-x^2+4x`

`quad=-1x^2+4x+0`

In this case, our `a` value is `-1`, the `b` value is 4, and the `c` value is `0`. This means the `x`-coordinate of the vertex is `(-4)/(2(-1))` which is `2`.

Now, to find the `y`-coordinate, simply plug `2` into the equation and see the value that it returns:

`y=-1x^2+4x+0`

`quad=>-1(2)^2+4(2)+0`

`quad=-1*4+8`

`quad=-4+8`

`quad=4`

Now we know that the vertex is at an `x` of `2` and a `y` of `4`, or the point `(2, 4)`.

You can check this by graphing the parabola: graph{4x-x^2 [-8, 12, -2, 8]}