1) What is the amplitude, period, horizontal shift (relative to basic function), and vertical translation of the following: a) -3cos(x+ 3.14/4) and b) 2sin(x-3.14/2)+3
2) evaluate the following expressions: a) log2 cubed root of 16 b) log25 (1/125) c) e^2ln5 d) ln 1/the cubed root of e^5
I will do 1 b) you do a)
y = 2 sin (x - π/2) + 3
amplitude = 2
period = 2π
phase shift = π/2 to the right
vertical translation : 3 units up
2. a) log216^(1/3)
= log2(2^4)^(1/3)
= log22^(4/3)
= (4/3) log2 2
= 4/3
b) log25 1/125
= log251 - log25125
= 0 - log25 25^(3/2)
= -3/2
c) e^2ln5
= e^(ln 5^2))
= 5^2 = 25 , based on property that e^lnk = k
d) ln (1/( e^5)^(1/3)
= ln1 - ln e^(5/3)
= 0 - 5/3 lne
= -5/3
1) For the function -3cos(x+ 3.14/4):
a) Amplitude:
The amplitude of a cosine function is the absolute value of the coefficient in front of the cosine term. In this case, the coefficient is -3, so the amplitude is 3 (since the amplitude is always positive).
b) Period:
The period of a cosine function is given by 2π divided by the coefficient in front of the x-term. In this case, the coefficient is 1, so the period is 2π.
c) Horizontal shift:
The horizontal shift is the value inside the parentheses after x. In this case, it is positive 3.14/4, which means the graph is shifted to the left by a quarter of a period.
d) Vertical translation:
There is no vertical translation in this function since there is no constant term added or subtracted.
For the function 2sin(x-3.14/2)+3:
a) Amplitude:
The coefficient in front of the sine term is 2, so the amplitude is 2.
b) Period:
The coefficient in front of the x-term is 1, so the period is 2π.
c) Horizontal shift:
The value inside the parentheses after x is -3.14/2, which means the graph is shifted to the right by half a period.
d) Vertical translation:
The constant term added at the end is 3, which means the graph is shifted upward by 3 units.
2) Evaluating the expressions:
a) log2(cubed root of 16):
First, find the cubed root of 16, which is 2. Then, take the logarithm base 2 of 2, which is equal to 1. So, log2(cubed root of 16) = 1.
b) log25(1/125):
Rewrite 1/125 as 5^(-3), since 125 is equal to (5^3). Then, take the logarithm base 25 of 5^(-3), which is equal to -3/2. So, log25(1/125) = -3/2.
c) e^(2ln5):
Using the logarithm properties, ln(a^b) = blog(e), we can rewrite the expression as e^(ln(5^2)), which simplifies to e^(2ln5). This is equivalent to (e^ln5)^2, which simplifies to 5^2. So, e^(2ln5) = 25.
d) ln(1/the cubed root of e^5):
First, evaluate the cubed root of e^5, which is e^(5/3). Then, take the natural logarithm of 1 divided by e^(5/3), which is equal to ln(1) - ln(e^(5/3)), or 0 - (5/3). So, ln(1/the cubed root of e^5) = -5/3.
1) To determine the amplitude, period, horizontal shift, and vertical translation of a trigonometric function, you can analyze the equation in the form:
a) -3cos(x + π/4)
Amplitude: The amplitude of a cosine function is the absolute value of the coefficient in front of the cosine term. In this case, the amplitude is |-3| = 3.
Period: The period of a cosine function is 2π divided by the coefficient of x. In this case, the coefficient of x is 1, so the period is 2π/1 = 2π.
Horizontal Shift: The horizontal shift is the opposite of the value inside the parentheses. In this case, the horizontal shift is -π/4.
Vertical Translation: There is no vertical translation in this function since there is no constant term added or subtracted.
b) 2sin(x - π/2) + 3
Amplitude: The amplitude of a sine function is the absolute value of the coefficient in front of the sine term. In this case, the amplitude is |2| = 2.
Period: The period of a sine function is 2π divided by the coefficient of x. In this case, the coefficient of x is 1, so the period is 2π/1 = 2π.
Horizontal Shift: The horizontal shift is the opposite of the value inside the parentheses. In this case, the horizontal shift is π/2.
Vertical Translation: The vertical translation is determined by the constant term added to the function. In this case, the function is shifted up 3 units.
2) Let's evaluate the given expressions:
a) log2(cube root of 16)
To evaluate this expression, first, find the cube root of 16.
Cube root of 16 = ∛16 = 2
Then, take the base 2 logarithm of the result.
log2(2) = 1
b) log25(1/125)
To evaluate this expression, rewrite the given fraction in terms of powers of 5.
1/125 = 5^(-3)
Then, take the base 25 logarithm of the result.
log25(5^(-3)) = -3
c) e^(2ln5)
To evaluate this expression, simplify the expression inside the parentheses first.
2ln5 = ln(5^2) = ln(25)
Then, raise e to the power of the result.
e^(ln(25)) = 25
d) ln(1/cube root of e^5)
To evaluate this expression, simplify the expression inside the parentheses first.
1/cube root of e^5 = 1/(e^(5/3))
Then, take the natural logarithm (base e) of the result.
ln(1/(e^(5/3))) = -(5/3)