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Next, some key points, like maximums, minimums, and intersections with the midline should be plotted. The parent sine function intersects the midline at each half-period.

In this case, the period is 660, so its half-period is 6602=330 nanometers. Therefore, the x-coordinates of the intersections of the function and the midline occur at x-values that are multiples of 330. x=0, 330, 660, ... These points lie on the midline, so their y-coordinate is 0.5.

The maximums and minimums of a sine function occur once every period between two points of intersection with the midline. Analyzing the graph of the parent sine function starting from the origin, it can be seen that the maximum of the function occurs before the minimum.

Therefore, the maximum of the given function is in the middle between the intersections (0,0.5) and (330,0.5), while the minimum is in the middle between (330,0.5) and (660,0.5). By adding and subtracting the amplitude of 0.3 to the midline, the y-coordinates of the maximum and minimum, respectively, can be found. cc Maximum& Minimum [0.2cm] (330/2,0.5 + 0.3) & (330 + 330/2,0.5 - 0.3) ⇕ & ⇕ (165,0.8) & (495,0.2) Now, plot both points on the coordinate plane.

Finally, connect the points with a smooth curve and continue it periodically.

Next, the key points should be identified and plotted. Consider the parent cosine function.

As can be seen, the maximums occur at x=- 2π, 0, 2π,..., which are the multiples of its period 2π. This means that using the period of 400 of the considered function, its maximums can be found. Maximums x=0, 400, 800, ... y=1.4 Furthermore, since the equation of the midline is y=0 and the amplitude is 1.4, the y-coordinate of the maximums is 1.4. The minimums of the parent function occur at x=- π, π,..., which are the values of its half-period. In this case, the half-period of the function is 200. Minimums x=200, 600, 1000, ... y=- 1.4 The y-coordinates of the minimums are - 1.4. Lastly, in-between every neighboring maximum and minimum are the intersections with the midline. Intersections x=100, 300, 500, ... y=0 Finally, plot all the found points on the coordinate plane with the midline.

By connecting the points with a smooth curve and continuing it periodically, the graph of the given function can be obtained.

External credits: @upklyak

d= 300-175tan x Therefore, to obtain the graph of d, first graph y=- 175 tan x and then translate it vertically. Start by comparing the function to the general form of a tangent function to identify the values of the coefficients a and b. General Form:& y= atan bx Given Function:& y= - 175 tan 1x Recall that the period of the parent tangent function is given by π|b|. Since b= 1, the period of the function can be calculated. Period π/| b| ⇒ π/| 1|=π One cycle of a tangent function occurs in the interval from - π2|b| to π2|b|. Substitute the value of b and find this interval for the given function. (- π/2| 1|,π/2| 1|) ⇔ (- π/2,π/2) Additionally, the asymptotes occur at the end of each cycle. This means that two asymptotes are x=- π2 and x= π2. Graph them on a coordinate plane.

Next, three points should be plotted between the asymptotes. By dividing the interval between - π2 and π2 into four equal parts, the x-coordinates of three points are obtained. x=- π/4, x=0, x=π/4 Substitute them into the function rule for x and find the corresponding y-coordinates.

Now, plot the points on the coordinate plane with the asymptotes.

Finally, the graph of y can be drawn by connecting the points with a smooth curve such that, as x tends to - π2 and π2, the graph gets closer and closer to the asymptotes but never crosses them.

The last step is to translate the graph 300 units up vertically to obtain the graph of d.

Therefore, only the part of the graph in the first quadrant makes sense considering the context.

Both functions have a general shape of a sine or a cosine function.

On the graph almost one full cycle of the function is shown and it can be assumed that it will repeat periodically moving to the right along the x-axis. By analyzing the graph's pattern, it can be concluded that for x=12, the function will have a value of 1250, the same value as for x=0. Therefore, the function has a period of 12 months. Period 12 months Additionally, the maximum value is 1750, while the minimum is 750. By calculating the mean of these values, the amplitude can be found. Amplitude [0.2cm] 1750- 750/2=500 Next, by subtracting the value of the amplitude from the maximum, or by adding its value to the minimum, the equation of the midline can be determined. Midline y=1250 Show all of these characteristics on the graph and try to identify which function, sine or cosine, can represent it.

Since almost one full cycle of the function is shown, it can be assumed that it will repeat periodically moving to the right along the x-axis. Also, it can be concluded that for x=12, the function will have the value 250, the same value as for x=0. This means that the function has a period of 12 months. Period 12 months The maximum value is y= 250, while the minimum is 150. Calculate the mean of these values to find the amplitude of the function. Amplitude [0.2cm] 250- 150/2=50 Now, by subtracting the value of the amplitude from the maximum, or by adding its value to the minimum, the equation of the midline can be written. Midline y=200 Show all of these characteristics on the graph and try to identify which function, sine or cosine, can represent it.

The diagram reminds the graph of a cosine function. Recall the general form of a cosine function. y=acos bm Here, a represents the amplitude and b represents the coefficient related to the period by the following formula. Period=2π/|b| Since for this function period is also 12 months, the value of |b| is π6. Taking into account only the positive values of a and b, substitute them into the equation. y= acos bm ⇓ y= 50cos π/6m Furthermore, the midline at y=200 indicates that the function has been translated 200 units up. This fact can be shown in the function rule by adding 200 to the right-hand side. f(m)=50cos π/6m+ 200 The function that models the population of foxes during the previous year was obtained using the graph and the values from the table, making it possible to determine the needed characteristics.

Similarly, by connecting the points given for the function f(m), its graph can be drawn.

The fox population seems to be behind the rabbit population by three months. In other words, the fox population chases the rabbit population. This can be explained by the predator-prey relationship that exists between foxes and rabbits, as rabbits are a source of food for foxes. Where there is a big population of rabbits, the fox population grows. Many rabbits → Foxes ↑ At the same time, when there are a lot of foxes, more rabbits are being hunted, so the rabbit population decreases. Many foxes → Rabbits ↓ When the rabbit population gets quite small, foxes have a limited food supply and, therefore, their population decreases. Few rabbits → Foxes ↓ Once the fox population is small enough, the rabbit population can recover, and this cycle continues. Few foxes → Rabbits ↑