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

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)#

#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!