Questions: Part II: Two Points You have been assigned two points. They are: (2,4) and (6,20) For these two points, you will need to: - Find a - sine and/or cosine model that contains the two points; - power function that contains the two points; - exponential function that contains the two points - For EACH Model provide a contextual interpretation, for example, x represents the number of seconds since the launch, and y represents the height of the spaceship. - For EACH model, - evaluate the model for some values of "x" and interpret the meaning of the result - Solve for the values of "x" for which the function "equals", "is greater than", or "is less than", some value, and interpret the meaning of the result - For EACH model, create a reasonable sketch that highlights its key features - For EACH model write a problem (contextually) that your model could be used to solve and then solve it.

Find a sine and/or cosine model that contains the two points \((2,4)\) and \((6,20)\).

Model Setup

A general sine or cosine model can be written as:
\[ y = A \sin(Bx + C) + D \quad \text{or} \quad y = A \cos(Bx + C) + D \]
We will use the sine model. The amplitude \(A\), period, phase shift \(C\), and vertical shift \(D\) need to be determined.

Determine Amplitude and Vertical Shift

The amplitude \(A\) is half the difference between the maximum and minimum values. The vertical shift \(D\) is the average of the maximum and minimum values. From the points \((2,4)\) and \((6,20)\), we assume the maximum is \(20\) and the minimum is \(4\). Thus:
\[ A = \frac{20 - 4}{2} = 8 \]
\[ D = \frac{20 + 4}{2} = 12 \]

Determine Period and Phase Shift

The distance between the two points is \(6 - 2 = 4\). Assuming this is half the period, the period \(T\) is \(8\). Thus:
\[ B = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \]
Using the point \((2,4)\):
\[ 4 = 8 \sin\left(\frac{\pi}{4} \cdot 2 + C\right) + 12 \]
\[ \sin\left(\frac{\pi}{2} + C\right) = -1 \]
\[ \frac{\pi}{2} + C = \frac{3\pi}{2} \]
\[ C = \pi \]

Final Model

The sine model is:
\[ y = 8 \sin\left(\frac{\pi}{4}x + \pi\right) + 12 \]

The sine model is \(\boxed{y = 8 \sin\left(\frac{\pi}{4}x + \pi\right) + 12}\).

Find a power function that contains the two points \((2,4)\) and \((6,20)\).

Model Setup

A power function can be written as:
\[ y = kx^n \]
We need to solve for \(k\) and \(n\) using the two points.

Solve for \(n\)

Using the points \((2,4)\) and \((6,20)\):
\[ \frac{20}{4} = \left(\frac{6}{2}\right)^n \]
\[ 5 = 3^n \]
\[ n = \log_3 5 \approx 1.465 \]

Solve for \(k\)

Using the point \((2,4)\):
\[ 4 = k \cdot 2^{1.465} \]
\[ k = \frac{4}{2^{1.465}} \approx 1.148 \]

Final Model

The power function is:
\[ y = 1.148x^{1.465} \]

The power function is \(\boxed{y = 1.148x^{1.465}}\).

Find an exponential function that contains the two points \((2,4)\) and \((6,20)\).

Model Setup

An exponential function can be written as:
\[ y = ab^x \]
We need to solve for \(a\) and \(b\) using the two points.

Solve for \(b\)

Using the points \((2,4)\) and \((6,20)\):
\[ \frac{20}{4} = \frac{ab^6}{ab^2} \]
\[ 5 = b^4 \]
\[ b = 5^{1/4} \approx 1.495 \]

Solve for \(a\)

Using the point \((2,4)\):
\[ 4 = a \cdot (1.495)^2 \]
\[ a = \frac{4}{(1.495)^2} \approx 1.792 \]

Final Model

The exponential function is:
\[ y = 1.792 \cdot (1.495)^x \]

The exponential function is \(\boxed{y = 1.792 \cdot (1.495)^x}\).

The sine model is \(\boxed{y = 8 \sin\left(\frac{\pi}{4}x + \pi\right) + 12}\).
The power function is \(\boxed{y = 1.148x^{1.465}}\).
The exponential function is \(\boxed{y = 1.792 \cdot (1.495)^x}\).