Question

[Trigonometric Graphs]

Use the following information to write an equation of the graph described:

10. sin; Amp = 4, per = 2π/3; phase shift = right π/4; vertical shift = up 3; reflect over x-axis.

12. cos; Amp = 2, period of 10π/3, reflect over x-axis.

Explain.​

Answer 1

Here's what I get.

Explanation:

Question 10

The general equation for a sine function is

y = a sin[b(x - h)] + k

Here's what the parameters control:

a = amplitude

k = vertical shift

b = the period (period = 2π/b; If period = 2π/3, b = 3)

h = horizontal shift

Reflect across y-axis (x ⟶ -x)

Your sine function will be:

`\begin{array}{rllll}y = & a \text{ sin}[ & b(x- & h)] + & k)\\& \downarrow & \downarrow & \downarrow & \downarrow\\& 4 & 3 & (\pi)/(4) & 3\\\end{array}`

Here are the effects of each parameter.

(a) Amp = 4

Increases the amplitude by a factor of 4 (Fig. 1)

(b) Up 3

k = 3. The graph shifts up three units (Fig. 2).

(c) Period = 2π/3

Set k = 3. The period changes from 2π to 2π/3 (Fig. 3).

Notice that you now hav3 three waves between 0 and 2π, where originally you had one.

(d) Right π/4.

Set h = 4. The graph shifts right by π/4.

notice how the trough at ½π shifts to ¾π.

(e) Reflect about y-axis

Set x equal to -x. Notice how the trough at (¾π, -1) is transformed to the trough at (-¾π, -1) (Fig. 5).

Question 12

y = a cos[b(x - h)] + k

a = 2

Per = 10π/3 ⟶ b = 3/5

h = 0

k = 0

Reflect over x-axis: y ⟶ -y

(a) a = 2

The amplitude is doubled.

(b) Per = 10π/3 ⟶ b = 3/5

The period (peak-to-peak distance) lengthens to 10π/3.

(c) Reflect about x-axis (y ⟶ -y)

All peaks are transformed into troughs and vice-versa.