102k views
4 votes
How do you graph j (x)=1/3x^2

User Andy Hume
by
7.5k points

2 Answers

3 votes
PART 1: Find the peak point (vertex) of the graph.
If the quadratic function form like this,
⇒ y = ax² + bx + c
the x-coordinate vertex is x = -b/2a

From the function above, we know that
⇒ a = 1/3
⇒ b = 0
⇒ c = 0
so the x-coordinate will be
x = -b/2a
x = 0/(2/3)
x = 0
Find the y-coordinate by plugging in the value of x to the function
y = 1/3(x²)
y = 1/3(0)
y = 0
The vertex = (0,0)

PART 2: Find two other points that pass through the graph
Plug in any x-coordinate to the function. For example, I put x = 3 and x = -3
For x = 3
y = (1/3)x²
y = (1/3)3²
y = (1/3)9
y = 3
The first point is (3,3)
For x = -3
y = (1/3)x²
y = (1/3)(-3)²
y = (1/3)9
y = 3
The second point is (-3,3)

PART C: Draw the graph
The point of coordinates we get from the function above are (0,0) as vertex, (3,3) and (-3,3). Put the coordinates to cartesian coordinate graph. (See the attachment)
How do you graph j (x)=1/3x^2-example-1
User Charles Young
by
7.6k points
1 vote
use desmos.com

or just plug in values for x
User Ichiban
by
7.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories