Question 1
Because the period is 2π, and the amplitude is 1obtain
f(x) = sin(x)
Because the horizontal shift is π, obtain
f(x) = sin(x - π)
Because the vertical shift is -4, obtain
f(x) = sin(x - π) - 4

Answer: 1. f(x) = sin(x - π) - 4

Question 2
The radius is 36/2 = 18 in.
1 revolution (360°) is the circumference, which is
2π(18) = 36π in
When the revolution is 62°, the distance traveled is
(62/360)*(36π) = (31/5)π in

Answer: 3. (31π)/5

Question 3.
Consider f(x) = 3cos(2x-π) - 1
f(0) = 3cos(-π) - 1 = -4
f(π/2) = 3cos(0) - 1 = 2
Rate of change = (2+4)/(π/2) = 12/π

From the graph, the rate of change of g(x) is
3/(π/2) = 6/π

Consider h(x) = sin(x) - 4
h(0) = 0 - 4 = -4
h(π/2) = 1 - 4 = -3
Rate of change = (-3+4)/(π/2) = 2/π
Therefore h(x) has the smallest rate of change

Answer: h(x)

8 0

3 years ago

1. Write the equation of the piece-wise function graphed below.

DaniilM [7]

Problem 4

<h3>Answer:</h3>

$f(x) = \begin{cases}2x+4 \text{ if } x < 1\\x-3 \text{ if } x \ge 1\end{cases}$

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Work Shown:

The left line goes through (-2,0) and (0,4)

The slope of this line is

m = (y2-y1)/(x2-x1)

m = (4-0)/(0-(-2))

m = (4-0)/(0+2)

m = 4/2

m = 2

The y intercept is b = 4

Since m = 2 and b = 4, this means y = mx+b turns into y = 2x+4. This portion is only done when x < 1. Note the open circle at the endpoint of this portion. So we do not include x = 1 as part of this piece.

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The line on the right side goes through (1,-2) and (2,-1)

Slope

m = (y2-y1)/(x2-x1)

m = (-1-(-2))/(2-1)

m = (-1+2)/(2-1)

m = 1/1

m = 1

The y intercept is b = -3. You can see this if you extend the line until it crosses the y axis.

Alternatively, plug in (x,y) = (1,-2) and m = 1 into y = mx+b to find that b = -3

So y = mx+b turns into y = 1x+(-3) or just y = x-3

We combine both parts to end up with $f(x) = \begin{cases}2x+4 \text{ if } x < 1\\x-3 \text{ if } x \ge 1\end{cases}$

This is only graphed when $x \ge 1$ (note the closed or filled in circle for the endpoint of this portion).

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Problem 5

Answer:

<h3> $f(x) = \frac{1}{2}|x+3|$ is the absolute value function</h3><h3>while this is the piecewise function</h3>

$f(x) = \begin{cases}-\frac{1}{2}(x+3) \text{ if } x < -3\\\frac{1}{2}(x+3) \text{ if } x \ge -3\end{cases}\\$

------------------

Work Shown:

y = |x| .... parent function

y = |x+3| ... shift 3 units to the left

y = (1/2)*|x+3| .... vertically compress by factor of 1/2

f(x) = (1/2)*|x+3|

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Break that down into a piecewise function

when x < -3, then y = -(1/2)(x+3)

when $x \ge -3$ , then y = (1/2)(x+3)

I'm using the rule that y = |x| turns into y = -x when x < 0 and y = x when $x \ge 0$

So that is how we get $f(x) = \begin{cases}-\frac{1}{2}(x+3) \text{ if } x < -3\\\frac{1}{2}(x+3) \text{ if } x \ge -3\end{cases}\\$ as the piecewise function.

8 0

3 years ago

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