Question

How do you find the vertex?

`y=x(x-2)`

Answer 1

See below

Explanation:

I assume you meant x(x-2)=0. Expressions like the one you wrote don't have vertices because they can't be graphed on a two-dimensional graph.
First, expand the expression on the left using the distributive property: `x^2-2x=0`

Now, you can write the expression in standard form: `ax^2+bx+c`. This looks like `x^2-2x+0=0`, so `a=1`, `b=-2`, and `c=0`.

To find the axis of symmetry, on which the vertex lies, use the equation `x=-b/(2a)`. When you substitute in the numbers from the equation, it's `x=-b/(2a) = -(-2)/(2*1) = 2/2 = 1`

To find the y-coordinate of the vertex, substitute the x-coordinate into the equation: `y=x^2-2x = (1)^2-(2*1) = 1 - 2 = -1`

So the vertex of `x(x-2) = 0` is `(1, -1)`.

Answer 2

The vertex is `(0,-2)`.

Explanation:

Given:

`y=x(x-2)`

Expand the right-hand side.

`y=x^2-2x`

This is a quadratic equation in standard form:

`y=ax^2+bx+c`,

where:

`a=1`, `b=-2`, `c=0`

To find the vertex of a quadratic equation in standard form, use the formula for the axis of symmetry to find the x-coordinate. Then substitute the `x`-coordinate into the equation and solve for `y` to get the `y`-coordinate.

`x=(-b)/(2a)`

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

`x=2/2`

`x=1`

Substitute `1` for `x` and solve for `y`.

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

`y=1-2`

`y=-1`

The vertex is `(1,-1)`.

graph{y=x^2-2x [-10, 10, -5, 5]}