Determine whether the quadratic function has a minimum or maximum value.Then find the coordinates of the minimum or maximum point.

Question

f(x)=2x^2-4x

Answer 1

if you know Calculus, then
f'(x) = 4x - 4
= 0 for a max/min
4(x-1) = 0
x = 1
then f(1) = 2-4 = -2
so (1,-2) is a min (since the parabola opens upwards)

if you don't know Calculus, use "completing the square" to find the vertex

f(x) = 2[x^2 - 2x + 1 - 1 ]
= 2[(x-1)^2 - 1]
= 2(x-1)^2 - 2
so the vertex is (1,-2) as above

or

you could use the formula:
the x of the vertex is -b/(2a)
= -(-4)/2(2)
= -1
etc

Answer 2

y = 2 (x^2-2x)
dy/dx = 2 (2x-2) = extreme at x = 1
d^2y/dx^2 = 4 so that extreme is a minimum
at x = 1
f(1) = -2