Question
Select the graph of the function y=(x−6)2−7
.(1 point)
Responses
A U-shaped function is graphed on the coordinate plane.
Image with alt text: A U-shaped function is graphed on the coordinate plane.
A U-shaped function is graphed on the coordinate plane.
Image with alt text: A U-shaped function is graphed on the coordinate plane.
A U-shaped function is graphed on the coordinate plane.
Image with alt text: A U-shaped function is graphed on the coordinate plane.
A U-shaped function is graphed on the coordinate plane.
Image with alt text: A U-shaped function is graphed on the coordinate plane.
Remaining Attempts : 3
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.(1 point)
Responses
A U-shaped function is graphed on the coordinate plane.
Image with alt text: A U-shaped function is graphed on the coordinate plane.
A U-shaped function is graphed on the coordinate plane.
Image with alt text: A U-shaped function is graphed on the coordinate plane.
A U-shaped function is graphed on the coordinate plane.
Image with alt text: A U-shaped function is graphed on the coordinate plane.
A U-shaped function is graphed on the coordinate plane.
Image with alt text: A U-shaped function is graphed on the coordinate plane.
Remaining Attempts : 3
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Answers
GPT-4o mini
To find the graph of the function \( y = (x - 6)^2 - 7 \), let's analyze its features based on the standard form of a quadratic function, which is \( y = a(x - h)^2 + k \).
1. **Vertex**: The vertex form of a quadratic function is given by \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. In your function:
- \( h = 6 \)
- \( k = -7 \)
This means the vertex of the graph is at the point \( (6, -7) \).
2. **Direction**: Since the coefficient \( a = 1 \) is positive, the parabola opens upwards.
3. **Y-intercept**: To find the y-intercept, set \( x = 0 \) in the equation:
\[
y = (0 - 6)^2 - 7 = 36 - 7 = 29
\]
So, the y-intercept is at the point \( (0, 29) \).
With the vertex at \( (6, -7) \), the parabola opens upward, and crosses the y-axis at \( (0, 29) \).
In order to select the correct graph, look for a U-shaped curve with its vertex at \( (6, -7) \), opening upwards, and ensure that it is consistent with the y-intercept calculation. If the graphs have the same characteristics as outlined, identify the one that matches these points.
1. **Vertex**: The vertex form of a quadratic function is given by \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. In your function:
- \( h = 6 \)
- \( k = -7 \)
This means the vertex of the graph is at the point \( (6, -7) \).
2. **Direction**: Since the coefficient \( a = 1 \) is positive, the parabola opens upwards.
3. **Y-intercept**: To find the y-intercept, set \( x = 0 \) in the equation:
\[
y = (0 - 6)^2 - 7 = 36 - 7 = 29
\]
So, the y-intercept is at the point \( (0, 29) \).
With the vertex at \( (6, -7) \), the parabola opens upward, and crosses the y-axis at \( (0, 29) \).
In order to select the correct graph, look for a U-shaped curve with its vertex at \( (6, -7) \), opening upwards, and ensure that it is consistent with the y-intercept calculation. If the graphs have the same characteristics as outlined, identify the one that matches these points.