can anyone help me?? f(x)= 2x^2?

Question

Find the Y intercept the eqyation of the axis of symmetry x coordinate of the vertex. Make a tabel of values that includes the vertex, and graph it??

Answer 1

The curve is a parabola.

To find y-intercept, put x=0
f(x) is zero for x=0 therefore the curve passes through origin and y-intercept is zero.

Changing x to -x, the value of f(x) remains unchanged. therefore the curve is symmetric about y-axis.
x 0 1 2 3 -1 -2 -3
f(x) 0 2 8 18 2 8 18

plot the points and you can draw the smooth curve.

you can have more points by giving different values to x.

Answer 2

The axis of symmetry is the y-axis; equation is x=0
The y-intercept is (0,0)
The vertex is at (0,0)
x=-1 -2 -3 0 1 2 3
y= 2 8 18 0 2 8 18

Answer 3

for quadratic functions in the form
f(x) = ax^2 + bx + c
the y-intercept is c, the axis of symmetry is x = -b/(2a) and that x value is also the x coordinate of the vertex.

For your problem, the y-intercept is 0 and the axis of symm is
x = 0. In fact the vertex is (0,0). So the table might look like:

x | y
------
-2 | 8
-1 | 2
0 | 0
1 | 2
2 | 8

Then plot those points and connect them with a curve (parabola)

Answer 4

the y intercept will be at 0, it will be a curve, and basically for every x square it then multiply it by 2, so if x=0 y=0, x=1 y=2,
x=-1 , y= 2 , and so on

Answer 5

F(x) = 0 when x=0, also F(0)= 0 so your x and y intercepts are (0,0) and (0,0) to make the table of values just plug in diferrent positive and negative numbers for x and solve to get a range of points to graph.

Answer 6

This is a graph of a parabola. (since x is squared)
The vertex is at the origin. V(0,0)
The y-intercept is at y=0.

Answer 7

x = 1 2 3 4 5
f(x) = 2 8 18 32 50

make a value for x and substitute to equation and you will get the answer. good luck!

Answer 8

y intercept is x=0
2(0)^2=0
x=0
vertex(0,0)
x=-2 y=8
x=-1 y=2
x=0 y=0
x=1 y=2
x=2 y=8