Question

How do you find the vertex for `f(x)=1/3x^2`?

Answer 1

`h=-b/(2a) = 0/(2*(1/3)) =0, k = f(0) = 0`
`vertex = (0,0)`

Explanation:

To find the x-coordinate of the vertex use the formula `h=-b/(2a)`. Note `a=1/3,b=0,c=0`. To find the y-coordinate k you take the x-value or h and put into the original equation for x and then solve.

Answer 2

vertex = (0,0)

Explanation:

The basic 'building block' for quadratic functions is y `= x^2`

The graph of this function has it's vertex at (0,0) and is symmetrical about the y-axis, as shown. graph{x^2 [-10, 10, -5, 5]}

for the general situation : y `= ax^2`

The graph shown above has a = 1.

If a > 1 the graph retains it's vertex at (0,0) and its shape but 'closes inwards in a similar way to a flower bud closing.

If a < 0 , the vertex is still (0,0) and shape is similar but this time it 'opens outward like the bud opening.
I've put `y = x^2" and " y = 1/3 x^2`on the same graph for you to compare them. The outer one is y`= 1/3x^2`
graph{(y-1/3 x^2)(y-x^2)=0 [-10, 10, -5, 5]}