Match each equation with the correct feature.(4 points)

Question

Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
y=−12(x−4)2+1
y=3(x−4)2−1
y=(x+1)2+4
y=(x+4)2−1
The option "Maximum at 1" (3 of 4) has been selected. Press tab to choose a response area, and spacebar to insert it. Press escape to cancel.

Answer 1

To match each equation with its correct feature, we need to analyze the given equations. These are all quadratic equations in vertex form, which is given by \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola.

1. \(y = -12(x-4)^2 + 1\):

   • This equation has a vertex at \((4, 1)\) and opens downwards since the coefficient of \((x-4)^2\) is negative.
   • Feature: Maximum at 1

2. \(y = 3(x-4)^2 - 1\):

   • This equation has a vertex at \((4, -1)\) and opens upwards since the coefficient of \((x-4)^2\) is positive.
   • Feature: Minimum at -1

3. \(y = (x+1)^2 + 4\):

   • This equation has a vertex at \((-1, 4)\) and opens upwards.
   • Feature: Minimum at 4

4. \(y = (x+4)^2 - 1\):

   • This equation has a vertex at \((-4, -1)\) and opens upwards.
   • Feature: Minimum at -1

Based on this analysis, here’s how they match up:

• \(y = -12(x-4)^2 + 1\): Maximum at 1
• \(y = 3(x-4)^2 - 1\): Minimum at -1
• \(y = (x+1)^2 + 4\): Minimum at 4
• \(y = (x+4)^2 - 1\): Minimum at -1

Please let me know if you need further assistance!