Question
the graph of a quadratic function is shown on the grid. which function is best represented by the graph?
the graph is upwards
a f(x) = x^2 + 12
b f(x) = 1/3x^2 + 12
c f(x) = -x^2 - x + 6
d f(x) = -1/3x^2 - x + 6
the graph is upwards
a f(x) = x^2 + 12
b f(x) = 1/3x^2 + 12
c f(x) = -x^2 - x + 6
d f(x) = -1/3x^2 - x + 6
Answers
GPT-4o mini
To determine which function best represents the graph of the quadratic function you described, we can analyze the characteristics of each option provided:
- **Option A: \( f(x) = x^2 + 12 \)**
- This is a standard quadratic function that opens upwards. The vertex will be at (0, 12), and it has a minimum point.
- **Option B: \( f(x) = \frac{1}{3}x^2 + 12 \)**
- This quadratic also opens upwards but is wider than the standard quadratic. The vertex will also be at (0, 12).
- **Option C: \( f(x) = -x^2 - x + 6 \)**
- This function is a downward-opening parabola (as indicated by the negative leading coefficient).
- **Option D: \( f(x) = -\frac{1}{3}x^2 - x + 6 \)**
- Similar to option C, this function is also a downward-opening parabola.
Since the graph is described as opening upwards, we can immediately eliminate options C and D. We are left with options A and B.
To determine which function might be more appropriate, we'd consider the vertex location. If the vertex is at (0, 12), then both options A and B could fit, but option B represents a wider parabola.
Without the graph for direct comparison, we can't definitively decide between A and B.
However, if you see that the parabola is not very narrow, and rises gently, then option B (\( f(x) = \frac{1}{3}x^2 + 12 \)) is more appropriate.
If it is steeper, then option A (\( f(x) = x^2 + 12 \)) is more appropriate.
Please use the visual features to make your final decision based on whether the parabola appears wider or steeper!
- **Option A: \( f(x) = x^2 + 12 \)**
- This is a standard quadratic function that opens upwards. The vertex will be at (0, 12), and it has a minimum point.
- **Option B: \( f(x) = \frac{1}{3}x^2 + 12 \)**
- This quadratic also opens upwards but is wider than the standard quadratic. The vertex will also be at (0, 12).
- **Option C: \( f(x) = -x^2 - x + 6 \)**
- This function is a downward-opening parabola (as indicated by the negative leading coefficient).
- **Option D: \( f(x) = -\frac{1}{3}x^2 - x + 6 \)**
- Similar to option C, this function is also a downward-opening parabola.
Since the graph is described as opening upwards, we can immediately eliminate options C and D. We are left with options A and B.
To determine which function might be more appropriate, we'd consider the vertex location. If the vertex is at (0, 12), then both options A and B could fit, but option B represents a wider parabola.
Without the graph for direct comparison, we can't definitively decide between A and B.
However, if you see that the parabola is not very narrow, and rises gently, then option B (\( f(x) = \frac{1}{3}x^2 + 12 \)) is more appropriate.
If it is steeper, then option A (\( f(x) = x^2 + 12 \)) is more appropriate.
Please use the visual features to make your final decision based on whether the parabola appears wider or steeper!