Question

move at least one of the 5 guide points below to complete the graph of $y = \frac{5}{2}x^2 - 2$. 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. graph with red and blue points, axes, and buttons: reset, reflect over x-axis

Explanation:

Step1: Analyze the vertex form

The function is \( y = \frac{5}{2}x^2 - 2 \), which is a quadratic function. The vertex form of a quadratic is \( y = a(x - h)^2 + k \), where \((h,k)\) is the vertex. Here, \( h = 0 \), \( k = -2 \), so the vertex (blue point) should be at \((0, -2)\). Currently, the blue point is at \((0,0)\), so we need to move it down 2 units to \((0, -2)\).

Step2: Analyze the red points (vertical stretch)

For a quadratic \( y = ax^2 + k \), when \( x = \pm 1 \), \( y = a(1)^2 + k=\frac{5}{2}(1) - 2=\frac{5}{2}-2=\frac{1}{2} \). When \( x = \pm 2 \), \( y=\frac{5}{2}(4)-2 = 10 - 2 = 8 \)? Wait, no, wait: \( y=\frac{5}{2}x^2 - 2 \), so for \( x = \pm 1 \), \( y=\frac{5}{2}(1)-2=\frac{5 - 4}{2}=\frac{1}{2} \approx 0.5 \). For \( x=\pm 2 \), \( y=\frac{5}{2}(4)-2 = 10 - 2 = 8 \)? Wait, no, the red points in the graph are at \( x=\pm 1 \) (lower red points) and \( x = \pm 2 \) (upper red points). Wait, the current lower red points (at \( x=\pm 1 \)) are at \( y = 1 \), but they should be at \( y=\frac{1}{2} \)? Wait, no, maybe I miscalculated. Wait, the original function is \( y=\frac{5}{2}x^2 - 2 \). Let's recalculate:

For \( x = 0 \): \( y = -2 \) (vertex).

For \( x = \pm 1 \): \( y=\frac{5}{2}(1)-2=\frac{5}{2}-2=\frac{1}{2}=0.5 \).

For \( x = \pm 2 \): \( y=\frac{5}{2}(4)-2 = 10 - 2 = 8 \). Wait, but in the graph, the upper red points are at \( y = 4 \) (when \( x=\pm 2 \)). So the vertical stretch factor: the standard parabola \( y = x^2 \) has points \((\pm 1,1)\), \((\pm 2,4)\). Our function has \( a=\frac{5}{2} \), so the \( y \)-values for \( x=\pm 1 \) should be \( \frac{5}{2}(1)=2.5 \) plus \( k=-2 \), so \( 0.5 \), and for \( x=\pm 2 \), \( \frac{5}{2}(4)=10 \) plus \( k=-2 \), so \( 8 \). So the red points at \( x=\pm 1 \) (lower red) should be at \( y = 0.5 \) (moved down from current \( y = 1 \)), and red points at \( x=\pm 2 \) (upper red) should be at \( y = 8 \) (moved up from current \( y = 4 \)). But the main point is the vertex (blue point) needs to be moved down to \((0, -2)\) to account for the vertical shift of \( -2 \), and the red points adjusted for the vertical stretch of \( \frac{5}{2} \).

So first, move the blue point (vertex) from \((0,0)\) to \((0, -2)\) (shift down 2 units). Then, for the red points: when \( x = \pm 1 \), the \( y \)-value should be \( \frac{5}{2}(1)^2 - 2=\frac{1}{2} \), so the lower red points (at \( x=\pm 1 \)) should be moved down from \( y = 1 \) to \( y = 0.5 \). For \( x=\pm 2 \), \( y=\frac{5}{2}(4)-2 = 8 \), so the upper red points (at \( x=\pm 2 \)) should be moved up from \( y = 4 \) to \( y = 8 \).

Answer:

To complete the graph:

1. Move the blue vertex point from \((0, 0)\) down to \((0, -2)\) (to account for the \( -2 \) vertical shift).
2. Adjust the red points: for \( x = \pm 1 \), move the lower red points to \( y=\frac{1}{2} \) (since \( y=\frac{5}{2}(1)^2 - 2=\frac{1}{2} \)), and for \( x = \pm 2 \), move the upper red points to \( y = 8 \) (since \( y=\frac{5}{2}(2)^2 - 2 = 8 \)) to account for the vertical stretch of \( \frac{5}{2} \).

(Note: The key initial step is moving the blue point to \((0, -2)\) to get the correct vertical shift, and then adjusting red points for the stretch. The final graph should have vertex at \((0, -2)\), and red points at \( (\pm 1, \frac{1}{2}) \) and \( (\pm 2, 8) \) (or adjusted based on the interface's guide points, but the main shift is the vertex down 2 units and red points adjusted for the stretch factor \( \frac{5}{2} \)).)