Question

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

Answer 1

Vertex `(1, -2)`

Explanation:

Given -

`y=2x^2-4x`

`x=(-b)/(2xxa)=(-(-4))/(2 xx 2)=4/4=1`

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

Vertex `(1, -2)`

Answer 2

The vertex is at `(1, -2)`.

Explanation:

`y = 2x^2 - 4x`

This quadratic equation is in standard form, or `y = color(red)(a)x^2 + color(blue)(b)x + color(magenta)(c)` (in this case there's no `c` because `c` is just zero)

We know that `color(red)(a = 2)` and `color(blue)(b = -4)`.

To find the vertex in a standard quadratic equation, we have to do two things.
`quadquad1.` Find the `x`-value of the vertex using the formula `x = (-color(blue)(b))/(2color(red)(a))`
`quadquadquadquadquadx = (-(color(blue)(-4)))/(2(color(red)(2)))`

`quadquadquadquadquadx = 4/4`

`quadquadquadquadquadx = 1`

`quadquad2.` Find the `y`-value of the vertex by plugging in our value for `quadquadquadquad` `x` back into the original equation
`quadquadquadquadquady = 2x^2 - 4x`

`quadquadquadquadquady = 2(1)^2 - 4(1)`

`quadquadquadquadquady = 2(1) - 4`

`quadquadquadquadquady = 2 - 4`

`quadquadquadquadquady = -2`

Therefore, our vertex is at `(1, -2)`. To check our answer, let's graph the equation:

(desmos.com)

As you can see, the vertex is indeed at `(1, -2)`.

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

Hope this helps!