k(x)=2x^2-4x?

Question

Graph the following quadratic function. Locate vertex and x-intercepts. The book completes the square and the comes up with the equation y=2(x-1)^2 - 2. But i came up with 2(x-1)^2 1

Answer 1

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

To find the x-intercepts, make K(x) = 0 and solve for x.

0 = 2x^2 - 4x
0 = 2x(x - 4)

Now, equate each factor to 0.

2x = 0
x - 4 = 0

This means x = {0, 4}.

To find the vertex, you must complete the square.

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

Factor out 2,

k(x) = 2(x^2 - 2x)

Add and subtract 1.

k(x) = 2(x^2 - 2x + 1) - 2
k(x) = 2(x - 1)^2 - 2

This means the vertex is at (1, -2).

Answer 2

I can't graph it unfortunately but substitute these coordinates to get a parabola: (0,0), (1,-2), (-1,6), (3,6), (-3,30), (5...
but the way to do this is to make x coordinates and substitute into the equation to get the y coordinates. As for the x intercepts are:
2x^2-4x
2x(x-2) =0
either x = 0 or x = 2
I tried my best but if it isn't enough, then I failed.
Dexter T

Answer 3

substituting y for k(x),
y = 2x^2 - 4x
y = 2(x^2 - 2x)
y = 2(x^2 - 2x + 1 - 1)
y = 2(x^2 - 2x + 1) - 2
y + 2 = 2(x - 1)^2
The vertex is at (1, - 2)
Going back to the original equation to find x-intercepts,
2x^2 - 4x = 0
x(x - 2) = 0
x = 0, 2

Answer 4

the answer is 4.3zx(1ysw- your ******* crazy if you think anyone is going to answer this.)

Answer 5

do your own homework

Answer 6

do YOUR OWN HOMEWORK!!!!!!