Question

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

The function \( y = x^2 - 5 \) is a parabola in the form \( y = ax^2 + k \), where \( a = 1 \) and \( k=-5 \). The vertex of the parabola \( y = x^2 \) is at \( (0,0) \), and for \( y = x^2 - 5 \), the vertex (the blue point) should be shifted down 5 units. So the vertex should be at \( (0, - 5) \).

Step2: Adjust the blue point

Move the blue point (vertex) from \( (0,0) \) to \( (0, - 5) \). For the red points, let's take a point like when \( x = 1 \), \( y=(1)^2 - 5=-4 \), and when \( x = - 1 \), \( y=(-1)^2 - 5=-4 \). The current red points at \( x = 1 \) and \( x=-1 \) have \( y = 1 \) (from the graph), so we need to move them down to \( y=-4 \). Similarly, for \( x = 2 \) and \( x = - 2 \), \( y=(2)^2 - 5 = - 1 \)? Wait, no, wait \( x = 2 \): \( y=4 - 5=-1 \)? Wait, no, \( x = 2 \): \( y = 2^2-5=4 - 5=-1 \)? Wait, the current red points are at \( x=\pm2 \) with \( y = 4 \). So we need to move the red points at \( x=\pm2 \) to \( y=-1 \)? Wait, no, let's recalculate. The function is \( y=x^2 - 5 \). So when \( x = 0 \), \( y=-5 \) (vertex). When \( x=\pm1 \), \( y=1 - 5=-4 \). When \( x=\pm2 \), \( y = 4-5=-1 \). When \( x=\pm3 \), \( y=9 - 5 = 4 \). Wait, maybe the red points are at \( x=\pm3 \)? Wait, the graph shows red points at \( x=\pm2 \) (wait, the x - axis has marks at - 2, - 1, 0, 1, 2,...). Wait, the red points are at \( x=\pm2 \) (the ones closer to the vertex) and \( x=\pm3 \)? Wait, no, the left red point is at \( x=-2 \) (since it's two units left of 0) and right red point at \( x = 2 \), and the other red points? Wait, maybe the key is to move the vertex (blue point) down to \( (0, - 5) \) and adjust the red points accordingly.

Answer:

Move the blue vertex point to \( (0, - 5) \) and adjust the red points (e.g., the red point at \( x = 1 \) should be moved to \( (1, - 4) \), at \( x=-1 \) to \( (-1, - 4) \), at \( x = 2 \) to \( (2, - 1) \), at \( x=-2 \) to \( (-2, - 1) \)) to match the function \( y=x^2 - 5 \). The main adjustment is moving the blue vertex down 5 units to \( (0, - 5) \).