Find Trigonometric Functions Given Their Graphs (Vertical Shift)

a) Finding Max and Min:
We know the standard range of the basic sine function is between -1 and 1:
\[ -1 \leq \sin(bx + c) \leq 1 \]

Multiply all parts by the amplitude \( a \) (assuming \( a > 0 \)):
\[ -a \leq a \sin(bx + c) \leq a \]

Add the vertical shift \( d \) to all parts:
\[ -a + d \leq a \sin(bx + c) + d \leq a + d \]

This tells us the absolute highest and lowest points of the function:
\[ \mathbf{\text{max} = d + a \quad \text{and} \quad \text{min} = d - a} \]

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b) Solving for \( a \) and \( d \):
Treat the equations from part (a) as a system of linear equations:
1) \( d + a = \text{max} \)
2) \( d - a = \text{min} \)

To find \( d \): Add the two equations together:
\( (d + a) + (d - a) = \text{max} + \text{min} \)
\( 2d = \text{max} + \text{min} \implies \mathbf{d = \dfrac{\text{max} + \text{min}}{2}} \)

To find \( a \): Subtract equation 2 from equation 1:
\( (d + a) - (d - a) = \text{max} - \text{min} \)
\( 2a = \text{max} - \text{min} \implies \mathbf{a = \dfrac{\text{max} - \text{min}}{2}} \)

1. Find Amplitude (\(a\)) and Vertical Shift (\(d\)):
Read the extreme values from the y-axis: \( \text{max} = 0 \) and \( \text{min} = -4 \).
\[ d = \dfrac{0 + (-4)}{2} = \mathbf{-2} \]
\[ a = \dfrac{0 - (-4)}{2} = \mathbf{2} \]

2. Find the Period (\(b\)):
Looking at the graph, one full cycle completes exactly at \( x = \dfrac{4\pi}{3} \).
Period \( P = \dfrac{4\pi}{3} \).
Using the formula: \( \dfrac{4\pi}{3} = \dfrac{2\pi}{b} \implies 4\pi b = 6\pi \implies \mathbf{b = \dfrac{3}{2}} \).

3. Find the Phase Shift (\(c\)) Algebraically:
Our partial equation is: \( y = 2 \sin\left( \dfrac{3}{2}x + c \right) - 2 \).
We need a clear point from the graph to plug in. At the y-intercept, the graph touches the peak: \( (0, 0) \).
Substitute \( x = 0 \) and \( y = 0 \):
\[ 0 = 2 \sin\left( \dfrac{3}{2}(0) + c \right) - 2 \]
\[ 2 = 2 \sin(c) \implies \sin(c) = 1 \]

The sine function equals 1 at \( \dfrac{\pi}{2} \) (plus any multiple of \( 2\pi \)). We choose the simplest positive value: \( \mathbf{c = \dfrac{\pi}{2}} \).

Conclusion:
\[ \mathbf{y = 2 \sin\left( \dfrac{3x}{2} + \dfrac{\pi}{2} \right) - 2} \]

1. Find Amplitude (\(a\)) and Vertical Shift (\(d\)):
From the graph, \( \text{max} = \dfrac{18}{5} \) and \( \text{min} = -\dfrac{6}{5} \).
\[ d = \dfrac{\frac{18}{5} + \left(-\frac{6}{5}\right)}{2} = \dfrac{\frac{12}{5}}{2} = \mathbf{\dfrac{6}{5}} \]
\[ a = \dfrac{\frac{18}{5} - \left(-\frac{6}{5}\right)}{2} = \dfrac{\frac{24}{5}}{2} = \mathbf{\dfrac{12}{5}} \]

2. Find the Period (\(b\)):
The given points \( \left( \dfrac{3}{5}, \dfrac{18}{5} \right) \) (a maximum) and \( \left( \dfrac{7}{5}, -\dfrac{6}{5} \right) \) (a minimum) represent exactly half of a wave cycle.
Half Period = \( \dfrac{7}{5} - \dfrac{3}{5} = \dfrac{4}{5} \).
Full Period \( P = 2 \times \dfrac{4}{5} = \dfrac{8}{5} \).
Using the formula: \( \dfrac{8}{5} = \dfrac{2\pi}{b} \implies 8b = 10\pi \implies \mathbf{b = \dfrac{5\pi}{4}} \).

3. Find the Phase Shift (\(c\)) Algebraically:
Our partial equation is: \( y = \dfrac{12}{5} \sin\left( \dfrac{5\pi}{4}x + c \right) + \dfrac{6}{5} \).
Substitute the maximum point \( x = \dfrac{3}{5}, y = \dfrac{18}{5} \):
\[ \dfrac{18}{5} = \dfrac{12}{5} \sin\left( \dfrac{5\pi}{4}\left(\dfrac{3}{5}\right) + c \right) + \dfrac{6}{5} \]
Subtract \( \dfrac{6}{5} \) from both sides:
\[ \dfrac{12}{5} = \dfrac{12}{5} \sin\left( \dfrac{3\pi}{4} + c \right) \]
Divide by \( \dfrac{12}{5} \):
\[ 1 = \sin\left( \dfrac{3\pi}{4} + c \right) \]

Sine equals 1 at \( \dfrac{\pi}{2} \). Therefore:
\[ \dfrac{3\pi}{4} + c = \dfrac{\pi}{2} \implies c = \dfrac{\pi}{2} - \dfrac{3\pi}{4} = \dfrac{2\pi}{4} - \dfrac{3\pi}{4} = \mathbf{-\dfrac{\pi}{4}} \]

Conclusion:
\[ \mathbf{y = \dfrac{12}{5} \sin\left( \dfrac{5\pi x}{4} - \dfrac{\pi}{4} \right) + \dfrac{6}{5}} \]

a) Finding the Period:
The parameter \( b \) inside the cosine is \( \dfrac{\pi}{12} \).
Period \( P = \dfrac{2\pi}{\pi/12} = 2\pi \times \dfrac{12}{\pi} = \mathbf{24 \text{ hours}} \).
(This makes perfect logical sense for a daily temperature cycle.)

b) Finding the Maximum Temperature:
The function is in the format \( a\cos(bt) + d \). We know \( \text{max} = d + |a| \).
Vertical shift \( d = 30 \) and amplitude \( |a| = |-8| = 8 \).
\( \text{max} = 30 + 8 = \mathbf{38^\circ C} \).

c) Evaluating Specific Times:
• 12 AM (\(t = 0\)):
\( T(0) = -8\cos(0) + 30 = -8(1) + 30 = \mathbf{22^\circ C} \)
• 6 AM (\(t = 6\)):
\( T(6) = -8\cos\left(\dfrac{6\pi}{12}\right) + 30 = -8\cos\left(\dfrac{\pi}{2}\right) + 30 = -8(0) + 30 = \mathbf{30^\circ C} \)
• 12 PM (\(t = 12\)):
\( T(12) = -8\cos\left(\dfrac{12\pi}{12}\right) + 30 = -8\cos(\pi) + 30 = -8(-1) + 30 = 8 + 30 = \mathbf{38^\circ C} \)
• 6 PM (\(t = 18\)):
\( T(18) = -8\cos\left(\dfrac{18\pi}{12}\right) + 30 = -8\cos\left(\dfrac{3\pi}{2}\right) + 30 = -8(0) + 30 = \mathbf{30^\circ C} \)

1. Find Amplitude (\(a\)) and Vertical Shift (\(d\)):
Max = 120, Min = 40.
\( d = \dfrac{120 + 40}{2} = 80 \)
\( a = \dfrac{120 - 40}{2} = 40 \)

2. Find the Period (\(b\)):
If it completes 5 bounces per second, its frequency is 5 Hz. The period is the time for one bounce.
Period \( P = \dfrac{1}{5} = 0.2 \) seconds.
\( 0.2 = \dfrac{2\pi}{b} \implies b = \dfrac{2\pi}{0.2} = 10\pi \).

3. Determine the Function and Phase Shift:
Because it starts at its maximum height at \( t = 0 \), it behaves perfectly like a positive cosine wave. No phase shift is needed (\( c = 0 \)).

Conclusion:
\[ \mathbf{h(t) = 40\cos(10\pi t) + 80} \]

1. Use the Origin Point:
The function passes through \( (0,0) \).
\( f(0) = A \sin(B \cdot 0) + D \)
\( 0 = A(0) + D \implies \mathbf{D = 0} \).
Because \( D = 0 \), the graph is not vertically shifted. The center line is the x-axis.

2. Use the Maximum Point:
The maximum value is 6. Because the center line is 0, the amplitude is exactly the distance from the center to the peak.
Therefore, \( \mathbf{A = 6} \).

3. Find the Period:
For an unshifted sine wave (\( D=0 \), no phase shift), the first maximum occurs at exactly one quarter of its period.
The peak happens at \( x = \dfrac{\pi}{4} \).
Therefore, \( \dfrac{1}{4} \times P = \dfrac{\pi}{4} \implies P = \pi \).
Using the period formula: \( \pi = \dfrac{2\pi}{B} \implies \mathbf{B = 2} \).

Conclusion:
\( A = 6, \quad B = 2, \quad D = 0 \)
The equation is \( \mathbf{f(x) = 6\sin(2x)} \).

1. Identify Constant Parameters:
The vertical shift \( D \) and the frequency multiplier \( B \) do not change between equivalent sine and cosine forms.
\( D = 5 \)
\( B = 4 \)

2. Handle the Amplitude Sign:
The original function has a negative amplitude (\( -3 \)). We need \( A > 0 \), so \( A = 3 \).
To absorb the negative sign, we can use the property \( -\sin(\theta) = \sin(\theta \pm \pi) \).
Let's add \( \pi \) to the inside:
\( y = 3 \sin\left(4x - \dfrac{\pi}{2} + \pi\right) + 5 \)
\( y = 3 \sin\left(4x + \dfrac{\pi}{2}\right) + 5 \)

3. Convert from Sine to Cosine:
Use the co-function identity: \( \sin(\theta) = \cos\left(\theta - \dfrac{\pi}{2}\right) \).
Substitute our entire inner expression as \( \theta \):
\( y = 3 \cos\left( \left(4x + \dfrac{\pi}{2}\right) - \dfrac{\pi}{2} \right) + 5 \)
The \( \dfrac{\pi}{2} \) and \( -\dfrac{\pi}{2} \) cancel each other out perfectly.

Conclusion:
The simplified equivalent function is: \[ \mathbf{y = 3 \cos(4x) + 5} \]