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

2 Answers
Bdub
Mar 5, 2016

#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.

Jim G.
Mar 5, 2016

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]}

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