Question

move at least one of the 5 guide points below to complete the graph of $y = \frac{5}{2}x^2$. moving the red points changes the vertical stretch or compression. moving the blue point shifts the function left/right/up/down. click the buttons below to start over or reflect over the x-axis. reset reflect over x-axis

Explanation:

Step1: Recall the function form

The function is \( y = \frac{5}{2}x^2 \), a quadratic function with vertex at \((0,0)\) and vertical stretch factor \(\frac{5}{2}\).

Step2: Analyze guide points

• For \( x = 1 \) (and \( x = -1 \)), calculate \( y \): \( y=\frac{5}{2}(1)^2=\frac{5}{2} = 2.5 \)? Wait, no, wait: Wait, the red points at \( x=\pm1 \) currently have \( y = 1 \) (from the graph, the lower red points near \( x=\pm1 \) have \( y\approx1 \)). But for \( y=\frac{5}{2}x^2 \), when \( x = 1 \), \( y=\frac{5}{2}(1)^2 = 2.5 \)? Wait, no, wait the upper red points: when \( x=\pm2 \), let's check. Wait, the upper red points are at \( x=\pm2 \) (since they are at \( x=-2 \) and \( x=2 \) approximately). For \( x = 2 \), \( y=\frac{5}{2}(2)^2=\frac{5}{2}\times4 = 10 \)? Wait, no, the current upper red points have \( y = 4 \). So we need to adjust the red points (vertical stretch) to match \( y=\frac{5}{2}x^2 \).

Step3: Adjust points

• The vertex (blue point) is at \((0,0)\), which is correct for \( y=\frac{5}{2}x^2 \) (vertex at origin).
• For \( x = 1 \) (and \( x = -1 \)): \( y=\frac{5}{2}(1)^2 = 2.5 \), so the lower red points (at \( x=\pm1 \)) should be moved up to \( y = 2.5 \).
• For \( x = 2 \) (and \( x = -2 \)): \( y=\frac{5}{2}(2)^2=\frac{5}{2}\times4 = 10 \), so the upper red points (at \( x=\pm2 \)) should be moved up to \( y = 10 \).

So, to complete the graph, move the red points at \( x=\pm1 \) up to \( y = \frac{5}{2}(1)^2 = 2.5 \) and the red points at \( x=\pm2 \) up to \( y=\frac{5}{2}(2)^2 = 10 \), while keeping the blue vertex at \((0,0)\).

Answer:

To complete the graph of \( y = \frac{5}{2}x^2 \):

• Keep the blue vertex (vertex of the parabola) at \((0,0)\) (no shift, since the function has no horizontal/vertical shift).
• For the red points at \( x = \pm1 \): Calculate \( y = \frac{5}{2}(1)^2 = \frac{5}{2} = 2.5 \), so move these lower red points up to \( y = 2.5 \).
• For the red points at \( x = \pm2 \): Calculate \( y = \frac{5}{2}(2)^2 = 10 \), so move these upper red points up to \( y = 10 \).

(Note: The key is adjusting the vertical stretch by moving the red points to match \( y = \frac{5}{2}x^2 \)’s \( y \)-values for \( x = \pm1, \pm2 \), and keeping the vertex at the origin.)