Question

graph the following function by moving the green and blue dots (if necessary).
$y = \frac{1}{3}x^2$

Explanation:

Step1: Recall the parent function

The parent function of a quadratic function is \( y = x^2 \), which has its vertex at \((0,0)\) and passes through points like \((1,1)\), \((2,4)\), \((3,9)\) etc.

Step2: Analyze the given function

The given function is \( y=\frac{1}{3}x^2 \). This is a vertical stretch or compression of the parent function. The coefficient \(\frac{1}{3}\) (between 0 and 1) means it is a vertical compression of the parent function \( y = x^2 \) by a factor of \(\frac{1}{3}\).

Step3: Find key points

• For \( x = 0 \), \( y=\frac{1}{3}(0)^2 = 0 \), so the vertex (green dot) is at \((0,0)\) (which is already correct in the graph).
• For \( x = 1 \), \( y=\frac{1}{3}(1)^2=\frac{1}{3}\approx0.33\)? Wait, no, wait the current graph has blue dots at \( x = \pm1 \) at \( y = 1 \)? Wait, no, let's recalculate. Wait, maybe the original graph is for \( y = x^2 \), but we need to adjust for \( y=\frac{1}{3}x^2 \). Wait, no, let's check the points:

Wait, if \( x = 3 \), \( y=\frac{1}{3}(3)^2=\frac{1}{3}\times9 = 3 \). Wait, the current blue dot at \( x = 3 \) is at \( y = 9 \), which is for \( y = x^2 \). So we need to adjust the blue dots.

Wait, let's find the correct y - values for the blue dots:

• When \( x=\pm1 \), \( y=\frac{1}{3}(1)^2=\frac{1}{3}\)
• When \( x = \pm2 \), \( y=\frac{1}{3}(4)=\frac{4}{3}\approx1.33 \)
• When \( x=\pm3 \), \( y=\frac{1}{3}(9) = 3 \)

But maybe the graphing tool has the blue dots at \( x=\pm1 \), \( x=\pm2 \), \( x=\pm3 \). So we need to move the blue dots down (since the coefficient is less than 1, compressing vertically).

Wait, the green dot is at the vertex \((0,0)\), which is correct. For the blue dots:

• At \( x = \pm1 \), instead of \( y = 1 \), it should be \( y=\frac{1}{3}\)
• At \( x=\pm2 \), instead of \( y = 4 \), it should be \( y=\frac{4}{3}\)
• At \( x=\pm3 \), instead of \( y = 9 \), it should be \( y = 3 \)

But since the problem says "by moving the green and blue dots (if necessary)", the vertex (green dot) is already at \((0,0)\) which is correct. For the blue dots, we need to adjust their y - coordinates according to \( y=\frac{1}{3}x^2 \).

For example, the blue dot at \( x = 3 \) (right side) should be at \( y = 3 \) (since \( \frac{1}{3}(3)^2=3 \)), the blue dot at \( x = 2 \) should be at \( y=\frac{4}{3}\approx1.33 \), and the blue dot at \( x = 1 \) should be at \( y=\frac{1}{3}\approx0.33 \). But maybe the graph is a bit different, but the key is to recognize that the function \( y=\frac{1}{3}x^2 \) is a vertical compression of \( y = x^2 \) by a factor of \( \frac{1}{3} \), so the vertex is at \((0,0)\) and the other points are scaled down by \( \frac{1}{3} \) in the y - direction.

Answer:

To graph \( y=\frac{1}{3}x^2 \):

• The green dot (vertex) remains at \((0,0)\) (correct as is).
• For the blue dots:
• At \( x=\pm1 \), move the blue dots to \( y = \frac{1}{3}\) (from \( y = 1 \) if it was for \( y = x^2 \)).
• At \( x=\pm2 \), move the blue dots to \( y=\frac{4}{3}\) (from \( y = 4 \) if it was for \( y = x^2 \)).
• At \( x=\pm3 \), move the blue dots to \( y = 3 \) (from \( y = 9 \) if it was for \( y = x^2 \)).

(Note: The process involves recognizing the vertical compression of the quadratic function and adjusting the y - coordinates of the blue dots accordingly while keeping the vertex (green dot) at \((0,0)\).)