# How do you find the vertex?

## $y = x \left(x - 2\right)$

Jun 6, 2018

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: ${x}^{2} - 2 x = 0$

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

To find the axis of symmetry, on which the vertex lies, use the equation $x = - \frac{b}{2 a}$. When you substitute in the numbers from the equation, it's $x = - \frac{b}{2 a} = - \frac{- 2}{2 \cdot 1} = \frac{2}{2} = 1$

To find the y-coordinate of the vertex, substitute the x-coordinate into the equation: $y = {x}^{2} - 2 x = {\left(1\right)}^{2} - \left(2 \cdot 1\right) = 1 - 2 = - 1$

So the vertex of $x \left(x - 2\right) = 0$ is $\left(1 , - 1\right)$.

Jun 6, 2018

The vertex is $\left(0 , - 2\right)$.

#### Explanation:

Given:

$y = x \left(x - 2\right)$

Expand the right-hand side.

$y = {x}^{2} - 2 x$

This is a quadratic equation in standard form:

$y = a {x}^{2} + b x + c$,

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.

$x = \frac{- b}{2 a}$

$x = \frac{- \left(- 2\right)}{2 \cdot 1}$

$x = \frac{2}{2}$

$x = 1$

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

$y = {1}^{2} - 2 \left(1\right)$

$y = 1 - 2$

$y = - 1$

The vertex is $\left(1 , - 1\right)$.

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