How do I find the vertex and the equation of the line of symmetry and graph? f(x)=1/8 x^2

Question

I just need to know how to solve the equation or get it to where it got 1/8x^2?

Answer 1

Given equation is f(x) = y = 1/8x^2

Compare it standard parabola equaton iis y = ax^2+bx+c

a = 1/8, b = 0, c = 0

Axis of symmetry is x = -b/2a = 0/2*1/8 = 0

To find vertex substitute x = 0 in gven equation.

y = 1/8(0)^2

y = 0

Vertex of f(x) = 1/8x^2 is origin.

Graph of y = 1/8x^2

Draw a coordinate plane.

plot the curve which is passes through origin.

Answer 2

The function is f ( x ) = y = (1/8)x ².

The above equation represent the parabola.

The standard form of parabola equation is (x - h)^2 = 4p (y - k), where (h, k) = vertex and p = directed distance from vertex to focus and axis of symmetry x = 0.

Write the equation : y = (1/8)x ² in standard form of parabola.

(x - 0)² = 8(y - 0)

(x - 0)² = 4(2)(y - 0).

Vertex : (h, k ) = (0, 0), and axis of symmetry : x = 0(y - axis).

Make a table : choose some values for x and find the corresponding values for y.

x	y = (1/8)x2	(x, y )
- 6	y = (1/8)(- 6)2 = (1/8)36 = 9/2	(- 6, 9/2)
- 4	y = (1/8)(- 4)2 = (1/8)16 = 2	(- 4, 2)
- 2	y = (1/8)(- 2)2 = (1/8)4 = 1/2	(- 2, 1/2)
0	y = (1/8)(0)2 = 0	(0, 0)
2	y = (1/8)(2)2 = (1/8)4 = 1/2	(2, 1/2)
4	y = (1/8)(4)2 = (1/8)16 = 2	(4, 2)
6	y = (1/8)(6)2 = (1/8)36 = 9/2	(6, 9/2)

Use these ordered pairs to graph the equation.

1.Draw a coordinate plane.

2.Plot the points.

3.Draw a smooth curve through these points.