Respuesta :
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
-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:
-3 + 1.3sec(x + 2) ⇒ 2 places to the left
-1 − 2sec(6x + 5) = -1 - 2sec6(x + 5/6) ⇒ 5/6 places to the left
3 + 0.5sec(2x + 1) = 3 + 0.5sec2(x + 1/2) ⇒ 1/2 places to the left
-3 + 2sec(4x + 2) = -3 + 2sec4(x + 1/2) ⇒ 1/2 places to the left
2 − 0.2sec(5x − 2) = 2 - 0.2sec5(x - 2/5) ⇒ 2/5 places to the right
-3 − 5sec(3x − 2) = -3 - 5sec3(x - 2/3) ⇒ 2/3 places to the right
-5 + 3sec(6x − 4) = -5 + 3sec6(x - 2/3) ⇒ 2/3 places to the right
-4 + 7sec(5x − 7) = -4 + 7sec5(x - 7/5) ⇒ 7/5 places to the right
Part B:
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
-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:
-5 + 3sec(6x − 4) ⇒ 5 places down
-4 + 7sec(5x − 7) ⇒ 4 places down
-3 − 5sec(3x − 2) ⇒ 3 places down
-3 + 2sec(4x + 2) ⇒ 3 places down
-3 + 1.3sec(x + 2) ⇒ 3 places down
-1 − 2sec(6x + 5) ⇒ 1 place down
2 − 0.2sec(5x − 2) ⇒ 2 places up
3 + 0.5sec(2x + 1) ⇒ 3 places up
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
-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:
-3 + 1.3sec(x + 2) ⇒ 2 places to the left
-1 − 2sec(6x + 5) = -1 - 2sec6(x + 5/6) ⇒ 5/6 places to the left
3 + 0.5sec(2x + 1) = 3 + 0.5sec2(x + 1/2) ⇒ 1/2 places to the left
-3 + 2sec(4x + 2) = -3 + 2sec4(x + 1/2) ⇒ 1/2 places to the left
2 − 0.2sec(5x − 2) = 2 - 0.2sec5(x - 2/5) ⇒ 2/5 places to the right
-3 − 5sec(3x − 2) = -3 - 5sec3(x - 2/3) ⇒ 2/3 places to the right
-5 + 3sec(6x − 4) = -5 + 3sec6(x - 2/3) ⇒ 2/3 places to the right
-4 + 7sec(5x − 7) = -4 + 7sec5(x - 7/5) ⇒ 7/5 places to the right
Part B:
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
-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:
-5 + 3sec(6x − 4) ⇒ 5 places down
-4 + 7sec(5x − 7) ⇒ 4 places down
-3 − 5sec(3x − 2) ⇒ 3 places down
-3 + 2sec(4x + 2) ⇒ 3 places down
-3 + 1.3sec(x + 2) ⇒ 3 places down
-1 − 2sec(6x + 5) ⇒ 1 place down
2 − 0.2sec(5x − 2) ⇒ 2 places up
3 + 0.5sec(2x + 1) ⇒ 3 places up