- The function is even, so its graph is symmetric with respect to the y-axis. This eliminates graphs B, E, H, and J.
- The function has a period of 2π, which eliminates graph I (period π).
- The function has a maximum value of 1 and a minimum value of -1, which eliminates graph G (maximum > 1).
- The function passes through the origin (0,0), which eliminates graphs A, C, D, and F (all have f(0) = 0, but we need to check other points).
- At x = π/2, the function should be 1. Graph A has f(π/2) = -1, so it's eliminated.
- At x = π/2, graph C has f(π/2) ≈ 0.5, not 1, so it's eliminated.
- At x = π/2, graph D has f(π/2) = -1, so it's eliminated.
- At x = π/2, graph F has f(π/2) = 0, so it's eliminated.
- The only remaining graph is C, but we already eliminated it. Let me recheck.
Wait, let's reconsider the properties. The function is sin(x). It is odd, not even. So my first step was wrong.
- The function is odd, so its graph is symmetric with respect to the origin. This eliminates graphs A, C, D, F, G, I (which are even or not odd).
- The function has a period of 2π, which eliminates graph I.
- The function has a maximum value of 1 and a minimum value of -1.
- The function passes through the origin (0,0).
- At x = π/2, f(π/2) = 1.
- At x = π, f(π) = 0.
- At x = 3π/2, f(3π/2) = -1.
Looking at the remaining graphs: B, E, H, J.
- Graph B: at x=π/2, it goes to infinity, not 1. Eliminated.
- Graph E: at x=π/2, it goes to negative infinity. Eliminated.
- Graph H: at x=π/2, it goes to positive infinity. Eliminated.
- Graph J: at x=π/2, it is 1; at x=π, it is 0; at x=3π/2, it is -1. It is odd, periodic with 2π, and ranges between -1 and 1.
Therefore, the correct graph is J.
Parent Tip: Review the logic above to help your child master the concept of trig graphs worksheet.
