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ArbitrLikvidat [17]
3 years ago
11

Arrange the transformations of the function y = sec x according to the resultant horizontal shifts, starting from left to right,

in the graph of the original secant function. If there are multiple transformations that cause the same horizontal shift, arrange them in ascending order beginning with the lowermost vertical shift. Tiles -3 + 2sec(4x + 2) -5 + 3sec(6x − 4) 3 + 0.5sec(2x + 1) -3 − 5sec(3x − 2) 2 − 0.2sec(5x − 2) -3 + 1.3sec(x + 2) -1 − 2sec(6x + 5) -4 + 7sec(5x − 7)
Mathematics
1 answer:
bagirrra123 [75]3 years ago
4 0
Part A:

Given a function f(x), the function f(x - a) gives the horizontal shift of f(x), a units to the right while the function f(x + a) gives the horizontal shift of f(x), a units to the left.

Given the functions

<span>-3 + 2sec(4x + 2)
-5 + 3sec(6x − 4)
3 + 0.5sec(2x + 1)
-3 − 5sec(3x − 2)
2 − 0.2sec(5x − 2)
-3 + 1.3sec(x + 2)
-1 − 2sec(6x + 5)
-4 + 7sec(5x − 7)

The arrangements of the transformations of the function y = sec x according to the resultant shifts, starting from left to right in the graph of the original secant function is as follows:

</span>-3 + 1.3sec(x + 2) ⇒ 2 places to the left
<span>-1 − 2sec(6x + 5) = -1 - 2sec6(x + 5/6) ⇒ 5/6 places to the left
</span><span>3 + 0.5sec(2x + 1) = 3 + 0.5sec2(x + 1/2) ⇒ 1/2 places to the left
</span><span>-3 + 2sec(4x + 2) = -3 + 2sec4(x + 1/2) ⇒ 1/2 places to the left
</span><span>2 − 0.2sec(5x − 2) = 2 - 0.2sec5(x - 2/5) ⇒ 2/5 places to the right
</span><span>-3 − 5sec(3x − 2) = -3 - 5sec3(x - 2/3) ⇒ 2/3 places to the right
</span>-5 + 3sec(6x − 4) = -5 + 3sec6(x - 2/3) ⇒ 2/3 places to the right
-4 + 7sec(5x − 7)<span> = -4 + 7sec5(x - 7/5) ⇒ 7/5 places to the right



Part B:
</span>
Given a function f(x), the function f(x) - b gives the vertical shift of f(x), b units down while the function f(x) + b gives the vertical shift of f(x), b units up.

Given the functions

<span>-3 + 2sec(4x + 2)
-5 + 3sec(6x − 4)
3 + 0.5sec(2x + 1)
-3 − 5sec(3x − 2)
2 − 0.2sec(5x − 2)
-3 + 1.3sec(x + 2)
-1 − 2sec(6x + 5)
-4 + 7sec(5x − 7)

If there are multiple transformations that cause the same horizontal shift, the arrangements of the transformations of the function y = sec x in ascending order beginning with the lowermost vertical shift in the graph of the original secant function is as follows:
</span>
<span>-5 + 3sec(6x − 4) ⇒ 5 places down
</span><span>-4 + 7sec(5x − 7) ⇒ 4 places down
-3 − 5sec(3x − 2) ⇒ 3 places down
</span><span>-3 + 2sec(4x + 2) ⇒ 3 places down
</span>-3 + 1.3sec(x + 2) ⇒ 3 places down
<span>-1 − 2sec(6x + 5) ⇒ 1 place down
</span>2 − 0.2sec(5x − 2) ⇒ 2 places up<span>
3 + 0.5sec(2x + 1) ⇒ 3 places up</span>
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