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son4ous [18]
3 years ago
10

Match transformation of the function y = csc x with description of the resultant shift in the original cosecant graph. Tiles -1

+ csc(x − π) -1 + csc(x + π) 1 + csc(x + π) 1 + csc(x − π) Pairs Resultant Shift in the Function's Graph Transformation of the Function The graph of csc x shifts one unit down and π radians to the left. arrowBoth The graph of csc x shifts one unit up and π radians to the right. arrowBoth The graph of csc x shifts one unit down and π radians to the right. arrowBoth The graph of csc x shifts one unit up and π radians to the left. arrowBoth
Mathematics
1 answer:
Advocard [28]3 years ago
5 0
We are given with the original trigonometric function
y = csc x
We are also given different transformations of the original trigonometric function
<span>-1 + csc(x − π)
-1 + csc(x + π)
1 + csc(x + π)
1 + csc(x − π) 
</span>An addition or subtraction in the domain will result to a vertical shift. If there is a 1 added to the function, the shift will be one unit upward and if there is a -1 added, the shift will be one unit downward.
In the argument of the cosecant function, addition or subtraction results to a horizontal shift.  The addition of π will result to a shift of π radians to the left and subtraction will result to shift in the opposite direction.
So, the answer are:
-1 + csc(x − π) - <span>The graph of csc x shifts one unit down and π radians to the right
</span>-1 + csc(x + π) - The graph of csc x shifts one unit down and π radians to the left
1 + csc(x + π) - The graph of csc x shifts one unit up and π radians to the left
1 + csc(x − π)  - The graph of csc x shifts one unit up and π radians to the right
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6 0
3 years ago
Read 2 more answers
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}

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

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.

---

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).

===================================================

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|

------

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
Read 2 more answers
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