Question

move at least one of the 5 guide points below to complete the graph of $y = 4x^2$. 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 \( y = 4x^2 \)

The parent function is \( y = x^2 \), and \( y = 4x^2 \) is a vertical stretch of the parent function by a factor of 4. For a quadratic function \( y = ax^2 \), the vertex is at \( (0,0) \) (the blue point here should be at \( (0,0) \), which it is, so we focus on the red points).

Step2: Find coordinates for \( y = 4x^2 \)

For \( x = 1 \) and \( x = -1 \): \( y = 4(1)^2 = 4 \) and \( y = 4(-1)^2 = 4 \). For \( x = 2 \) and \( x = -2 \): \( y = 4(2)^2 = 16 \)? Wait, no, wait the graph's y - axis goes up to 10? Wait, maybe the red points are at \( x=\pm1 \) and \( x = \pm2 \)? Wait, looking at the current red points: the inner red points (closer to the vertex) are at \( x=-1 \) and \( x = 1 \), current y - value seems to be 1 (since \( y\approx1 \)). But for \( y = 4x^2 \), when \( x = 1 \), \( y = 4 \), and when \( x = 2 \), \( y = 16 \) (but the graph's y - axis only goes up to 10, maybe the red points are at \( x=\pm1 \) and \( x=\pm2 \) with correct y - values. Wait, the problem says "move at least one of the 5 guide points". Let's take the red point at \( x = 1 \) (the inner right red point). Currently, its y - value is 1 (from the graph, since it's near y = 1). We need to move it to \( y = 4(1)^2=4 \). Similarly, the red point at \( x=-1 \) (inner left) should be moved to \( y = 4 \). The outer red points (at \( x = 2 \) and \( x=-2 \)): for \( y = 4x^2 \), when \( x = 2 \), \( y = 16 \), but the graph's y - axis only goes up to 10, maybe the outer red points are at \( x=\pm1 \)? Wait, maybe the initial red points: the two inner red points (x = - 1, x = 1) have y = 1, and the two outer red points (x=-2, x = 2) have y = 4. But for \( y = 4x^2 \), when x = 1, y = 4; when x = 2, y = 16 (too big for the graph). Wait, maybe the graph is scaled, but the key is to move the red points to their correct y - values for \( y = 4x^2 \). So, for example, take the red point at (1,1) (inner right) and move it up to (1,4), and the red point at (- 1,1) (inner left) up to (-1,4). The outer red points (at x = - 2 and x = 2, current y = 4) should be moved up to y = 16, but since the graph's y - axis is up to 10, maybe there's a miscalculation. Wait, maybe the function is \( y = 4x^2 \) with the graph's y - axis adjusted, but the main action is to move the red points (vertical stretch points) to their correct y - values based on \( y = 4x^2 \). So, move the inner red points (x = ±1) up to y = 4 (since \( 4(1)^2 = 4 \)) and the outer red points (x = ±2) up to y = 16 (but if the graph can't go that high, maybe the problem has a typo, but the process is to use the function \( y = 4x^2 \) to find the correct y - coordinates for the x - values of the red points and move them accordingly.

Answer:

To complete the graph of \( y = 4x^2 \):

1. For the red points at \( x = 1 \) and \( x=-1 \) (the inner red points), move them vertically to \( y = 4 \) (since \( y = 4x^2 \) gives \( y = 4(1)^2=4 \) and \( y = 4(-1)^2 = 4 \)).
2. For the red points at \( x = 2 \) and \( x=-2 \) (the outer red points), move them vertically to \( y = 16 \) (since \( y = 4(2)^2=16 \) and \( y = 4(-2)^2 = 16 \)) if the graph's y - axis range allows. If not, adjust based on the graph's scale, but the key is to use the function \( y = 4x^2 \) to determine the correct y - coordinates for the x - values of the red guide points and move them to those positions.