News
Get Premium
Dashboard

Dashboard Sign out
Core Connections Algebra 2, 2013
CC
Core Connections Algebra 2, 2013 View details
Chapter Closure
search

Exercise 189 Page 435

a Since the sine and cosine curve are horizontal translations of each other, we are free to pick either function to model the graph. Examining the curve, we can determine its period, amplitude, and midline.
The curve has a period of 2π, an amplitude of 1, and a midline of 0. This fits both the description of a cosine curve and of a sine curve. However, since the graph intercepts the y-axis at (1,0), we know it actually describes the cosine curve perfectly.

y=cos x Let's check the curve on our graphing calculator. To draw a graph on a calculator, we first push the Y= button and type the function in one of the rows. Having written the function, we can push GRAPH to draw it.

As we can see, the graph on the calculator matches the graph from the book.

b Like in Part A, we will identify the amplitude, period, and midline of the graph.
The curve has a period of 2π, an amplitude of 1, and a midline of - 3. With this information, we know that the curve has been vertically translated by 3 units down. Compared with a sine and a cosine curve, we notice that it looks like an inverted sine curve. This means the coefficient to sin x must be -1.

y=- sin x-3 Let's check the curve on our graphing calculator.

As we can see, the graph on the calculator matches the graph from the exercise.

c Like in previous parts, we begin by identifying the period, amplitude, and midline of the curve.
The curve has a period of π2, an amplitude of 5, and a midline of 0. To determine the coefficient to x, we should equate the period with 2πb and solve for b.
period=2π/b
Substitute

period= π/2

π/2=2π/b
▼
Solve for b
MultEqn

LHS * b=RHS* b

bπ/2=2π
MultEqn

LHS * 2=RHS* 2

bπ=4π
DivEqn

.LHS /π.=.RHS /π.

b=4
Now we have to choose between a sine and a cosine curve. Since the sine curve intercepts the y-axis at (0,0) it is easier to use a sine curve, as we then will not need to horizontally translate the curve. y=5sin4x Let's check the curve on our graphing calculator.

As we can see, the graph on the calculator matches the graph from the exercise.

d Like in previous parts, we begin by identifying the period, amplitude, and midline of the graph.
The curve has a period of 6π, an amplitude of 1, and a midline of 0. To determine the coefficient to x, we should equate the period with 2πb and solve for b.
period=2π/b
Substitute

period= 6π

6π=2π/b
▼
Solve for b
MultEqn

LHS * b=RHS* b

6π b=2π
DivEqn

.LHS /6π.=.RHS /6π.

b=2π/6π
ReduceFrac

a/b=.a /2π./.b /2π.

b=1/3
Now we have to choose between a sine and a cosine curve. Since the sine curve intercepts the y-axis at (0,0) the graph is closer to a sine curve then a cosine curve. y=sin1/3x However, the given graph does not intercept the y-axis at the origin like a sine curve does. It has been horizontally translated to the right by π2 units. To reflect this horizontal translation, we have to subtract π2 from x. y=sin 1/3(x-π/2) Let's check the curve on our graphing calculator.

As we can see, the graph on the calculator matches the graph from the exercise.

Subchapter links expand_more
arrow_right Exercises
178
Exercises 179 (Page 433)
180
Exercises 181 (Page 433)
Exercises 182 (Page 434)
Exercises 183 (Page 434)
Exercises 184 (Page 434)
185
Exercises 186 (Page 434)
Exercises 187 (Page 434)
188
Exercises 189 (Page 435)
Exercises 190 (Page 435)