Answer
See the figure below.
Work Step by Step
We know that the position of the object that undergoes a simple harmonic motion is given by
$$x_{(t)}=A\cos(\omega t+\phi_0)$$
where $A$ is the amplitude which is 8 cm, $\omega=2\pi f$ where $f$ is the frequency and its value is 0.25 Hz, $\phi_0$ is the phase constant and it is $\phi_0=-\dfrac{\pi}{2}=270^\circ $ and since it is negative so the object is moving to the right.
Hence,
$$x_{(t)}=8\cos(2\pi f t+\dfrac{-\pi}{2} )$$
$$x_{(t)}=8\cos(\dfrac{\pi}{2} t-\dfrac{ \pi}{2} )$$
We know that the period is given by $T=1/f=1/0.25=4$ s, so the initial position at $t=0$ is $x=0$ cm.
We need to show 2 cycles of the motion, so $t_i=0$ s and $t_f=2T=8$ s.
See the figure below.