Graphical Reasoning Involving Trigonometric Functions

[nycmathguy]

Section 4.5

How we do find the value of a and d from the given graphs? How about doing 65 explaining each step along the way?

Thanks

 

[MathLover1]

65.
f(x)=a*cos(x)+d

a is amplitude=> the graph goes between horizontal lines y=-1 and y=3, max = 3 and min = - 1
a = (max - min) / 2 = 2
a = (3 - (-1)) / 2 = 4/2
a=2

your function is shifted (from y=-1) 2 units up , so, d=1

f(x)=2*cos(x)+1

 

[nycmathguy]

65.
f(x)=a*cos(x)+d

a is amplitude=> the graph goes between horizontal lines y=-1 and y=3, max = 3 and min = - 1
a = (max - min) / 2 = 2
a = (3 - (-1)) / 2 = 4/2
a=2

your function is shifted (from y=-1) 2 units up , so, d=1

f(x)=2*cos(x)+1

You said:

your function is shifted (from y=-1) 2 units up , so, d=1

From y = -1 to y = 3, that is a jump of 4 units not 2 units. Right? I will try the other three and post my work here this weekend.

 

[nycmathguy]

Question 66

This is of the form f(x) = a•cos(x) + d.

The letter a is the amplitude.

The graph lies between y = -3 and y = 1.

Max = 1; min = -3

a = (max -min)/2

a = 2

The function is shifted up 2 units from y = -3 and y = -1. This means d = -1.

I came up with the function f(x) = 2cos(x) - 1.

You say?

 

[MathLover1]

correct

 

[nycmathguy]

correct

This type of question, in my opinion, is more interesting than simply graphing a function or a group of functions. Agree?

 

[MathLover1]

Agree

 

[nycmathguy]

Agree

Cool. Not in the mood to do anything today but I will try to get with the program. My roommate is driving me up the wall.