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 May 19th, 2017, 08:57 PM #1 Newbie   Joined: May 2017 From: North America Posts: 7 Thanks: 0 Function Transformation I got my homework back and these are the questions that absolutely puzzled me. I really need help in understanding 1.Given the function f(x)=x+6. The function g(x) results after the function f(x) has been stretched about the x axis by a factor of 3, reflected in the x axis, translated 2 units to the right, and 4 units down. Determine the equation of g(x) in the simplest form. 3.The point (3, -4) is on the graph of y=f(x). Determine a point that would be on the graph of y=f(-(x+2))-1 5.The domain of y=f(x) is [2,6]. Determine the domain of y=1/2f(-2(x-1))+3. 7.The point (6,4) is on the graph of y=f(x) is transformed to become y=g(x). The point (3,-2) is on the graph of y=g(x) after certain transformations have occurred. List a set of possible transformation, in the correct order that may have occurred. 8. Given f(x)=x^2 and g(x) = f^-1(x). List all the invariant points when f(x) is transformed into g(X). Last edited by SN2; May 19th, 2017 at 08:59 PM. Reason: Error
June 2nd, 2017, 05:39 AM   #2
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Quote:
 Originally Posted by SN2 I got my homework back and these are the questions that absolutely puzzled me. I really need help in understanding 1.Given the function f(x)=x+6. The function g(x) results after the function f(x) has been stretched about the x axis by a factor of 3, reflected in the x axis, translated 2 units to the right, and 4 units down. Determine the equation of g(x) in the simplest form.
"Stretching along the x axis by a factor of 3" changes x to 3x which means that the "new" x is 1/3 the "old x" so f(x)= x/3+ 6. "Reflecting about the x-axis" swaps x and -x so f(x)= -x/3+ 6. "Translating 2 units to the right" changes x to x+ 2 so f(x)= -(x+ 2)/3+ 6. "Translating 4 units down" changes y to y- 4 so f(x)- 4= -(x+ 2)/3+ 6. Then f(x)= -(x+ 2)/3+ 10.

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 3.The point (3, -4) is on the graph of y=f(x). Determine a point that would be on the graph of y=f(-(x+2))-1.
If -(x+2)= 3 then x+ 2= -3 and x= -5. If f(3)= -4 then y= -4- 1= -5. The point is (-5, -5).

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 5.The domain of y=f(x) is [2,6]. Determine the domain of y=1/2f(-2(x-1))+3.
You must have -2(x- 1) between 2 and 6: 2<= -2(x-1)<= 6. Dividing by the negative number -2 reverses the direction of the inequality so -3<= x- 1<= -1 and then -2<= x<= 0. The domain of the new function is [-1, 0].

Quote:
 7.The point (6,4) is on the graph of y=f(x) is transformed to become y=g(x). The point (3,-2) is on the graph of y=g(x) after certain transformations have occurred. List a set of possible transformation, in the correct order that may have occurred.
The x value of 6 changes to 3 so one possible transformation is that x changes to 2x (so that 2*3 is 6). The y value of 4 changes to -2 so another possible transformation, after the first, is that y changes to y/-2. Another possible transformation for this is that y changes to y- 6.

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 8. Given f(x)=x^2 and g(x) = f^-1(x). List all the invariant points when f(x) is transformed into g(X).
An inverse function swaps x and y. To be invariant under that transformation, we must have y= x. For f(x)= y= x^2, we must have x= x^2 so x^2- x= x(x- 1)= 0. The "invariant points" are (0, 0) and (1, 1).

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