Question

graph $h(x) = 2\sin(\pi x - 3\pi) - 4$ in the interactive widget. note that one moveable point always defines an extremum point in the graph and the other point always defines a neighbouring intersection with the midline.

Explanation:

Step1: Simplify the sine function

Use sine subtraction identity:
$\sin(\pi x - 3\pi) = \sin(\pi x)\cos(3\pi) - \cos(\pi x)\sin(3\pi)$
Since $\cos(3\pi)=-1$, $\sin(3\pi)=0$, we get:
$\sin(\pi x - 3\pi) = -\sin(\pi x)$
So $h(x) = -2\sin(\pi x) - 4$

Step2: Identify key parameters

Amplitude: $|A|=2$, Midline: $y=-4$, Period: $T=\frac{2\pi}{\pi}=2$

Step3: Find extremum (maximum)

Max of $-2\sin(\pi x)-4$ occurs when $\sin(\pi x)=-1$.
$\pi x = \frac{3\pi}{2} + 2k\pi \implies x = \frac{3}{2} + 2k$. For $k=-2$, $x=-2$
$h(-2) = -2\sin(-2\pi)-4 = 0 - 4 + 2 = -2$, so point $(-2, -2)$

Step4: Find midline intersection

Midline $y=-4$, set $h(x)=-4$:
$-2\sin(\pi x)-4=-4 \implies \sin(\pi x)=0$
$\pi x = k\pi \implies x=k$. For $k=0$, $x=0$, point $(0, -4)$

Answer:

Extremum point: $(-2, -2)$, Midline intersection point: $(0, -4)$