Solution:

Given: $y = 2\sin(x) + 1$

Compare to general form: $y = A\sin(B(x - C)) + D$

• $A = 2$
• $B = 1$
• $C = 0$
• $D = 1$

Features:

1. Amplitude: $|A| = |2| = 2$
2. Period: $\frac{2\pi}{|B|} = \frac{2\pi}{1} = 2\pi$
3. Phase Shift: $C = 0$ (no horizontal shift)
4. Midline: $y = D = 1$
5. Maximum: $1 + 2 = 3$
6. Minimum: $1 - 2 = -1$

Key Points (one period from $x = 0$ to $x = 2\pi$):

• Start: $(0, 1)$ (midline)
• Max: $(\frac{\pi}{2}, 3)$
• Midline: $(\pi, 1)$
• Min: $(\frac{3\pi}{2}, -1)$
• End: $(2\pi, 1)$ (midline)

The graph oscillates between $y = -1$ and $y = 3$, centered on the midline $y = 1$.

Solution:

The general form is $f(x) = A\sin(B(x - C)) + D$ or $f(x) = A\sin(Bx - BC) + D$

Our function: $f(x) = 3\sin\left(2x - \frac{\pi}{3}\right) + 1$

Rewrite in standard form: $f(x) = 3\sin\left(2\left(x - \frac{\pi}{6}\right)\right) + 1$

Part (a): Amplitude $= |A| = |3| = 3$

Part (b): Period $= \frac{2\pi}{|B|} = \frac{2\pi}{2} = \pi$

Part (c): Phase shift $= C = \frac{\pi}{6}$ (to the right)

Alternatively, from $2x - \frac{\pi}{3} = 0$, we get $x = \frac{\pi}{6}$

Part (d): Vertical shift $= D = 1$

The midline is $y = 1$.

Solution:

General form: $y = A\cos(B(x - C)) + D$

Given information:

• Amplitude: $|A| = 3$, so $A = 3$ (assuming positive)
• Period: $\frac{2\pi}{B} = \pi$
• Phase shift right: $C = \frac{\pi}{4}$
• Vertical shift down: $D = -2$

Find $B$: $\frac{2\pi}{B} = \pi$ $2\pi = \pi B$ $B = 2$

Equation: $y = 3\cos(2(x - \frac{\pi}{4})) - 2$

Or simplified: $y = 3\cos(2x - \frac{\pi}{2}) - 2$

Verification:

• Amplitude: $|3| = 3$ ✓
• Period: $\frac{2\pi}{2} = \pi$ ✓
• Phase shift: $\frac{\pi}{4}$ right ✓
• Vertical shift: $-2$ ✓
• Range: $[-2-3, -2+3] = [-5, 1]$

Solution:

Part (a): Rewrite: $g(x) = -2\cos\left(\frac{\pi}{4}(x + 4)\right) - 3$

Here: $A = -2$, $B = \frac{\pi}{4}$, $C = -4$, $D = -3$

• Amplitude $= |A| = |-2| = 2$
• Period $= \frac{2\pi}{|B|} = \frac{2\pi}{\pi/4} = 8$
• Phase shift $= C = -4$ (4 units to the left)
• Midline: $y = D = -3$

Part (b): For cosine, the range is $[-1, 1]$.

With amplitude 2 and reflection: $-2\cos(...) \in [-2, 2]$

Adding vertical shift of $-3$:

Maximum: $2 + (-3) = -1$ Minimum: $-2 + (-3) = -5$

Part (c): Set $g(x) = 0$:

$-2\cos\left(\frac{\pi x}{4} + \pi\right) - 3 = 0$

$-2\cos\left(\frac{\pi x}{4} + \pi\right) = 3$

$\cos\left(\frac{\pi x}{4} + \pi\right) = -\frac{3}{2}$

But wait! The range of cosine is $[-1, 1]$, and $-\frac{3}{2} < -1$.

Therefore, there are no $x$-intercepts (the graph never crosses the $x$-axis).

This makes sense because the maximum value is $-1$, which is still below the $x$-axis.

Solution:

Given: $y = -\frac{1}{2}\tan(\frac{x}{2}) + 3$

Compare to general form: $y = A\tan(B(x - C)) + D$

• $A = -\frac{1}{2}$ (negative means reflection over x-axis)
• $B = \frac{1}{2}$
• $C = 0$
• $D = 3$

Features:

1. Period: $\frac{\pi}{|B|} = \frac{\pi}{1/2} = 2\pi$
2. Vertical shift: $y = 3$ (midline)
3. Reflection: Negative $A$ means graph is reflected over the midline
4. Vertical asymptotes occur at $x = \frac{\pi}{2B} + \frac{n\pi}{B}$

Find asymptotes: $x = \frac{\pi}{2(1/2)} + \frac{n\pi}{1/2} = \pi + 2n\pi$

For $n = 0, 1, 2$:

• $n = 0$: $x = \pi$
• $n = 1$: $x = 3\pi$
• $n = 2$: $x = 5\pi$ (outside interval)

In $[0, 4\pi]$, asymptotes are at $x = \pi$ and $x = 3\pi$.

Key points between asymptotes:

Between $x = -\pi$ and $x = \pi$:

• $(0, 3)$ (midline, where tangent crosses zero)
• $(-\frac{\pi}{2}, 3.5)$ (quarter period from center)
• $(\frac{\pi}{2}, 2.5)$ (quarter period from center)

The graph decreases from top to bottom (due to negative $A$) within each period, with the center line at $y = 3$.