0/3 attempts next attempt points: 10? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 sketch a graph of the…

Explanation:

Step1: Identify the general form

The function is of the form $y = A\sin(Bx - C)+D$, where for $f(x)=- 2\sin(\pi x)+3$, $A=-2$, $B = \pi$, $C = 0$, $D = 3$.

Step2: Determine the amplitude

The amplitude $|A|=| - 2|=2$. This means the graph oscillates between $y = 3 + 2=5$ and $y=3 - 2 = 1$.

Step3: Find the period

The period $T=\frac{2\pi}{B}$. Since $B=\pi$, $T=\frac{2\pi}{\pi}=2$.

Step4: Locate key - points

For a sine function $y = \sin x$, key - points are at $x = 0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi$. For $y=-2\sin(\pi x)+3$, when $x = 0$, $y=-2\sin(0)+3=3$; when $x=\frac{1}{2}$, $y=-2\sin(\frac{\pi}{2})+3=-2 + 3=1$; when $x = 1$, $y=-2\sin(\pi)+3=3$; when $x=\frac{3}{2}$, $y=-2\sin(\frac{3\pi}{2})+3=2 + 3=5$; when $x = 2$, $y=-2\sin(2\pi)+3=3$.

Step5: Sketch the graph

Plot the key - points $(0,3),(\frac{1}{2},1),(1,3),(\frac{3}{2},5),(2,3)$ and connect them with a smooth sine - like curve. The graph is a sine curve with an amplitude of 2, period of 2, reflected about the $x$ - axis (due to negative $A$) and shifted up 3 units.

Answer:

Sketch a sine - like curve with amplitude 2, period 2, reflected about the $x$ - axis and shifted up 3 units, passing through key - points $(0,3),(\frac{1}{2},1),(1,3),(\frac{3}{2},5),(2,3)$.