graph the given function for - 2π ≤ x ≤ 2π, adjusting the vertical window as needed. state whether or not…

Explanation:

Step1: Analyze the components of the function

The function $f(x)=x^{2}-2\sin x$ is a combination of a quadratic function $y = x^{2}$ (which is a parabola opening upwards with vertex at $(0,0)$) and a sine - function $y=- 2\sin x$. The sine function $y = \sin x$ has an amplitude of 1, and $y=-2\sin x$ has an amplitude of 2.

Step2: Evaluate the function at key points

When $x = 0$, $f(0)=0^{2}-2\sin(0)=0$. When $x=\pm2\pi$, $f(\pm2\pi)=(2\pi)^{2}-2\sin(2\pi)=(2\pi)^{2}\approx(2\times3.14)^{2}=39.44$. The quadratic term $x^{2}$ dominates as $|x|$ increases. The sine function $y = - 2\sin x$ oscillates between - 2 and 2, but its effect is relatively small compared to the growth of $x^{2}$ as $|x|$ gets large.

Step3: Determine periodicity

A function $y = f(x)$ is periodic if there exists a non - zero constant $T$ such that $f(x + T)=f(x)$ for all $x$ in the domain. Let's assume $f(x + T)=(x + T)^{2}-2\sin(x + T)=x^{2}-2\sin x=f(x)$. Expanding $(x + T)^{2}-2\sin(x + T)=x^{2}+2Tx+T^{2}-2\sin(x + T)$. The quadratic term $x^{2}$ will not satisfy the periodicity condition for non - zero $T$ since of the $2Tx+T^{2}$ terms. So the function $f(x)=x^{2}-2\sin x$ is not periodic. The graph of $y = x^{2}-2\sin x$ is a parabola - like curve with some wiggles due to the sine term. As $x$ moves away from 0 in either the positive or negative direction, the $x^{2}$ term causes the function to increase.

Answer:

The function is not periodic. Without seeing the actual graphs A, B, C in detail, we know that the graph should be a parabola - like curve opening upwards with some small oscillations due to the sine term. If we had to choose based on the general shape and non - periodicity, we would look for a graph that has a parabolic shape and does not repeat itself over the interval $[-2\pi,2\pi]$.