How do you graph j (x)=1/3x^2
2 answers:
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)
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)
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