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

Mar 5, 2016

$h = - \frac{b}{2 a} = \frac{0}{2 \cdot \left(\frac{1}{3}\right)} = 0 , k = f \left(0\right) = 0$
$v e r t e x = \left(0 , 0\right)$

#### Explanation:

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

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 $= a {x}^{2}$

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 $y = {x}^{2} \text{ and } y = \frac{1}{3} {x}^{2}$on the same graph for you to compare them. The outer one is y$= \frac{1}{3} {x}^{2}$
graph{(y-1/3 x^2)(y-x^2)=0 [-10, 10, -5, 5]}