Question

move at least one of the 5 guide points below to complete the graph of $y = 7x^2 - 4$. 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: Identify the vertex form of the parabola

The given function is \( y = 7x^2 - 4 \), which is a quadratic function in the form \( y = ax^2 + k \). The vertex of this parabola is at \( (0, -4) \) because for \( y = ax^2 + k \), the vertex is \( (0, k) \). Currently, the blue point (vertex) is at \( (0, 0) \), so we need to move the blue point down to \( (0, -4) \).

Step2: Determine the y - values for red points

For the red points, let's take \( x = 1 \) and \( x=- 1 \) (the inner red points) first. Substitute \( x = 1 \) into \( y=7x^2 - 4 \): \( y=7(1)^2-4=7 - 4 = 3 \). Substitute \( x = 2 \) and \( x = - 2 \) (the outer red points) into \( y = 7x^2-4 \): \( y=7(2)^2-4=7\times4 - 4=28 - 4 = 24 \)? Wait, no, looking at the graph, the outer red points are at \( x=\pm2 \)? Wait, no, in the graph, the outer red points are at \( x = \pm2 \)? Wait, the current outer red points are at \( y = 4 \) when \( x=\pm2 \)? Wait, no, let's re - evaluate. The function is \( y = 7x^2-4 \). Let's take \( x = 1 \): \( y=7(1)^2-4 = 3 \), \( x = 2 \): \( y=7(4)-4=24 \)? But the graph's y - axis only goes up to 10. Wait, maybe the red points are at \( x=\pm1 \) and \( x=\pm2 \) with different y - values. Wait, the initial red points (inner) are at \( x=\pm1 \), current y - value is 1. We need to change their y - value to \( y = 7(1)^2-4=3 \). The outer red points (at \( x=\pm2 \)) have current y - value 4, and we need to change it to \( y=7(2)^2 - 4=28 - 4 = 24 \)? But the graph's y - axis is up to 10, so maybe the x - values of the red points are \( x=\pm1 \) and \( x=\pm\sqrt{\frac{y + 4}{7}} \). Wait, the key is to move the blue point (vertex) to \( (0,-4) \) and adjust the red points. The blue point (vertex) is the one that shifts the graph up/down. So first, move the blue point from \( (0,0) \) to \( (0, - 4) \). Then, for the red points at \( x=\pm1 \), calculate \( y=7(1)^2-4 = 3 \), so move the inner red points (at \( x = \pm1 \)) up to \( y = 3 \). For the red points at \( x=\pm2 \), calculate \( y=7(2)^2-4=24 \), but since the graph's y - axis is up to 10, maybe the x - values of the outer red points are \( x=\pm1 \)? No, the initial graph has two pairs of red points: one pair at \( x=\pm1 \) (inner) and one pair at \( x=\pm2 \) (outer). So to complete the graph, we move the blue vertex to \( (0, - 4) \), move the inner red points (at \( x=\pm1 \)) to \( (1, 3) \) and \( (-1, 3) \), and move the outer red points (at \( x=\pm2 \)) to \( (2, 24) \)? But since the y - axis is up to 10, maybe there's a misinterpretation. Wait, the problem says "Move at least one of the 5 guide points". The most important point to move first is the blue vertex (the vertex of the parabola) from \( (0,0) \) to \( (0, - 4) \) to account for the vertical shift of - 4, and then adjust the red points for the vertical stretch (since \( a = 7>1 \), it's a vertical stretch).

Answer:

To complete the graph of \( y = 7x^2-4 \):

1. Move the blue vertex (the point at the origin \( (0,0) \)) down to \( (0, - 4) \) to account for the \( - 4 \) vertical shift.
2. For the red points at \( x=\pm1 \) (inner red points), move them up to \( y = 7(1)^2-4=3 \) (so the points are \( (1, 3) \) and \( (-1, 3) \)) to account for the vertical stretch by a factor of 7.
3. For the red points at \( x=\pm2 \) (outer red points), move them up to \( y=7(2)^2-4 = 24 \) (but if the graph's y - axis range is limited, adjust according to the graph's scale, but the main action is to move the vertex and the red points to match the function \( y = 7x^2-4 \)). The key initial move is moving the blue vertex to \( (0, - 4) \).