# How do you find the vertex of y=2x^2-4x?

May 6, 2018

Vertex $\left(1 , - 2\right)$

#### Explanation:

Given -

$y = 2 {x}^{2} - 4 x$

$x = \frac{- b}{2 \times a} = \frac{- \left(- 4\right)}{2 \times 2} = \frac{4}{4} = 1$

At x=1; y=2(1^2)-4(1)=2-4=-2

Vertex $\left(1 , - 2\right)$ May 6, 2018

The vertex is at $\left(1 , - 2\right)$.

#### Explanation:

$y = 2 {x}^{2} - 4 x$

This quadratic equation is in standard form, or $y = \textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{m a \ge n t a}{c}$ (in this case there's no $c$ because $c$ is just zero)

We know that $\textcolor{red}{a = 2}$ and $\textcolor{b l u e}{b = - 4}$.

To find the vertex in a standard quadratic equation, we have to do two things.
$\quad \quad 1.$ Find the $x$-value of the vertex using the formula $x = \frac{- \textcolor{b l u e}{b}}{2 \textcolor{red}{a}}$
$\quad \quad \quad \quad \quad x = \frac{- \left(\textcolor{b l u e}{- 4}\right)}{2 \left(\textcolor{red}{2}\right)}$

$\quad \quad \quad \quad \quad x = \frac{4}{4}$

$\quad \quad \quad \quad \quad x = 1$

$\quad \quad 2.$ Find the $y$-value of the vertex by plugging in our value for $\quad \quad \quad \quad$ $x$ back into the original equation
$\quad \quad \quad \quad \quad y = 2 {x}^{2} - 4 x$

$\quad \quad \quad \quad \quad y = 2 {\left(1\right)}^{2} - 4 \left(1\right)$

$\quad \quad \quad \quad \quad y = 2 \left(1\right) - 4$

$\quad \quad \quad \quad \quad y = 2 - 4$

$\quad \quad \quad \quad \quad y = - 2$

Therefore, our vertex is at $\left(1 , - 2\right)$. To check our answer, let's graph the equation: (desmos.com)

As you can see, the vertex is indeed at $\left(1 , - 2\right)$.

For more help with finding the vertex from the standard equation, feel free to watch this video:

Hope this helps!