Question

Write the equation of a sine or cosine function to describe the graph.

Answer 1

Step-by-step explanation:

The equation of a trigonometric function is

$y = a \sin(bx + c) + d$

or

$y = a \: cos(bx + c) + d$

Let define some variables,

D is the midline, this refers to the midpoint of the highest y value and lowest y value. Some textbooks call it the vertical translations but it is the same thing.

A is the amplitude. The amplitude is the distance from the midline to the highest y value. Some distance is non negative, the formula for the amplitude is

$|a|$

The period is how often the wave repeats itself on a interval.

Period can't be negative so the formula for period is

$\frac{2 \pi}{ |b| }$

To find the period, look at the extreme points.

The phase shift tells us if the sinusoid have been shifted to the right or left. The formula for the phase shift

$bx + c = 0$

Solving for x gives us

$x = \frac{ - c}{b}$

If our x is negative, we have a phase shift to the right

If our x is Positve, we have a phase shift to the left.

Let solve this equation, Let use Sine since sin(0)=0,

The smallest y value here is 0, and the highest is 2, so the amplitude is 1.

$d = 1$

Next, the distance from the max to the midline is 1, as well the min to the midline is also 1.

So

$a = 1$

We have minimum at 0, and 8, so our period is 8.

$\frac{2\pi}{b} = 8$

Solve for b,

$\frac{\pi}{4} = b$

Plug this in the equation.

$1 \sin( \frac{\pi}{4} x) +$

$y = \sin( \frac{\pi}{4} x) + 1$

The graph passes through (4,2) so let see if that holds true for our equation

$2 = \sin( \frac{\pi}{4} (4)) + 1$

$2 = \sin(\pi) + 1$

$2 = 0 + 1$

$2 = 1$

This doesn't hold true, so we must have a phase shift

Notice that

$\sin( \frac{\pi}{2} ) = 1$

So if we shift this pi/2 to the right, we can get our equation to be true.

$y = \sin( \frac{\pi}{4} x - \frac{\pi}{2} ) + 1$

This equation works so our equation is

$y = \sin( \frac{\pi}{4} x - \frac{\pi}{2} ) + 1$