Question

Graph a sine function whose amplitude is 4, period is π , midline is y=−3, and y-intercept is (0, −3). The graph is not a reflection of the parent function over the x-axis.

Use the sine tool to graph the function.

The first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.

Answer 1

y=4sin(2x) - 3

Step-by-step explanation:

Recall the form of a sinusoidal function is Asin(Bx + C) + D, where A is the amplitude, B is the frequency (could also be viewed as the period), C is the horizontal (phase) shift, and D is the vertical shift. Also, recall sin(x) has a period of 2π.

You start with the parent function (sin(x)).

Then, knowing the amplitude is 4: 4sin(Bx + C) + D

Then, knowing the period of the regular sine function is 2π, to get to a period of π, divide by 2 (the period of the curve changes by a factor of 1/2), so your B coefficient is 2: 4sin(2x + C) + D

There is no phase shift, so omit C: 4sin(2x) + D

Since the midline is y = -3, that is your vertical shift (D): 4sin(2x) - 3

Graph of y = 4sin(2x) - 3 is below:

Answer 2

Answer: The graph is shown below.

Step-by-step explanation:

The midline is y = -3 and the y intercept is (0,-3). This is point A in the diagram below.

The period is pi units, which means that going from midline to peak (or valley) is 1/4 of this distance. So it's pi*(1/4) = pi/4 = 0.79 units approximately from x = 0 to x = 0.79; which is the x coordinate of point B. Point B is one of the infinitely many maximum points. Notice that the vertical distance from A to B is 4 units, which is the amplitude.

Extra info:

The equation of this sine curve is y = 4sin(2x)-3

I used GeoGebra to make the graph