To determine which graph shows a function where \( f(2) = 4 \), we need to find the graph where the point \( (2, 4) \) lies on the curve.
Let's analyze each graph description:
1. **Graph 1**: A curved line with an upward arc crosses the y-axis at \( (0, 1) \) and exits the plane at points \( (-4, 14) \) and \( (4, 14) \).
- This graph has a lowest point (vertex) within the range (between -4 and 4). The curvature seems to indicate a parabola opening upwards. We need to check if \( (2, 4) \) could feasibly be on this curve. Given that several points on this arc reach values like 14 at the endpoints, it is possible that this curve passes through \( (2, 4) \).
2. **Graph 2**: A curved line with a downward arc crosses the y-axis at \( (0, -1) \) and exits at points \( (-4, -14) \) and \( (4, -14) \).
- This graph forms a parabola opening downwards. Given that \( f(2) = 4 \) requires the value at \( x = 2 \) to be positive, and this graph only reaches negative values at those x-coordinates, we can rule out this graph.
3. **Graph 3**: A curved line with an upward arc exits the plane at points \( (7, 14) \) and \( (13, 14) \) with the bottom of the arc at point \( (10, 3) \).
- This graph is unlikely to include \( x = 2 \) given that its domain (based on the described points) starts at \( x = 7 \) and goes to \( x = 13 \). Hence, \( (2, 4) \) does not lie on this curve.
4. **Graph 4**: A curved line with an upward arc exits the plane at points \( (-14, 12) \) and \( (-6, 14) \) with the bottom of the arc at point \( (-10, -3) \).
- Similar to the previous one, this graph's described points are all in the negative x-region. This makes it impossible for \( (2, 4) \) to be on this curve.
Based on the given descriptions, the correct graph where \( f(2) = 4 \) lies on the function is the **first graph** because it is the only graph where the point \( (2, 4) \) could reasonably fit within the span of the upward arc.
Which graph shows a function where f(2) = 4?
On a coordinate plane, a curved line with an upward arc crosses the y-axis at (0, 1) and exits the plane at points (negative 4, 14) and (4, 14).
On a coordinate plane, a curved line with a downward arc crosses the y-axis at (0, negative 1) and exits at points (negative 4, negative 14) and (4, negative 14).
On a coordinate plane, a curved line with an upward arc exits the plane at points (7, 14) and (13, 14) with the bottom of the arc at point (10, 3).
On a coordinate plane, a curved line with an upward arc exits the plane at points (negative 14, 12) and (negative 6, 14) with the bottom of the arc at point (negative 10, negative 3).
1 answer