Math Transformations Of A Function

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Jun 16th 2009, 10:36 PM
AlphaRock

Math Transformations Of A Function

#1 If the point (10,6) is on the graph of y = f(x), what point is on the graph of y = f(-2x-4)? Answer: (-7,6)

My work:
y = f(-2(x+2))
... Unsure how to complete this and go on.

#2 The zeros of the function y = f(x) are -4, 1, and 2. Determine the zeros of the new function y = -f(x-1).

My work:
(-4,0)
(1,0)
(2,0)

-4+1 = -3
-1+1 = 0
2+1 = 3
(-3) x -1 = 3
0 x -1 = 0
3 x -1 = -3

#3 The maximum value of y = f(x) is 6. Find the maximum value of y = (1/3)f((1/2)x).

My work: (1/3)f((1/2)(6)) = 1. Not right though.
Jun 16th 2009, 11:47 PM
yeongil

This should have been posted in the Pre-Calculus subforum. (Wink)

Quote:

Originally Posted by AlphaRock

#3 The maximum value of y = f(x) is 6. Find the maximum value of y = (1/3)f((1/2)x).

My work: (1/3)f((1/2)(6)) = 1. Not right though.

Don't plug in 6 for x!! 6 is a y value.

Let's compare $f(x)$ to $f\left(\frac{1}{2}x\right)$ . The latter is a horizontal stretch by a factor of 2. The minimum and maximum y values, if any, do not change. The maximum value of the latter function would still be y = 6.

(Look at a concrete example: $y = \sin x$ and $y = \sin\left(\frac{1}{2}x\right)$ . In both cases, the minimum and maximum values are -1 and 1, respectively.)

Okay, now let's compare $f\left(\frac{1}{2}x\right)$ to $\frac{1}{3}f\left(\frac{1}{2}x\right)$ . This time, the latter has a vertical shrink by a factor of 1/3. The minimum and maximum y values, if any, this time, will change. The maximum value of the latter function is now y = 2 (just multiply 6 by 1/3).

(Look at another concrete example: $y = \sin x$ vs. $y = \frac{1}{3}\sin x$ . The min & max values for the original function are -1 and 1, but the min & max values for the transformed function are -1/3 and 1/3.)

Does that help?

01
Jun 16th 2009, 11:57 PM
yeongil

Quote:

Originally Posted by AlphaRock

#2 The zeros of the function y = f(x) are -4, 1, and 2. Determine the zeros of the new function y = -f(x-1).

My work:
(-4,0)
(1,0)
(2,0)

-4+1 = -3
-1+1 = 0
2+1 = 3
(-3) x -1 = 3
0 x -1 = 0
3 x -1 = -3

Don't do the red -- it's wrong (even though you have the same numbers (-3, 0, & 3) as you had before multiplying by -1).

You are correct to add 1 to each of the zeroes of the function, because when you go from $f(x)$ to $f(x - 1)$ you have a horizontal shift to the right by 1 unit.

However, when you go from $f(x - 1)$ to $-f(x - 1)$ , you have reflection on the x-axis. The zeroes of the function do not change. (If you have a graphing calculator, try graphing $y = x^2 - 6$ and $y = -(x^2 - 6)$ , and you'll see that the x-intercepts for both graphs are the same.) What is happening is that the y-values are being multiplied by -1, not the x-values. And at the x-intercepts, 0 times -1 is still 0.

So the x-intercepts for $-f(x - 1)$ are -3, 0, and 3.

01
Jun 19th 2009, 09:58 AM
HallsofIvy

Quote:

Originally Posted by AlphaRock

#1 If the point (10,6) is on the graph of y = f(x), what point is on the graph of y = f(-2x-4)? Answer: (-7,6)

My work:
y = f(-2(x+2))
... Unsure how to complete this and go on.

You only know f(10) so in order to be able to calculate f(-2x-4), you must have -2x-4= 10. Solve that for x. And, of course, the only y value you know is "6" so if -2x-4= 10, f(-2x-4)= f(10)= 6.

Quote:

#2 The zeros of the function y = f(x) are -4, 1, and 2. Determine the zeros of the new function y = -f(x-1).
(-4,0)
(1,0)
(2,0)

My work:
-4+1 = -3
-1+1 = 0
2+1 = 3

You know that f(-4)= 0, f(1)= 0, and f(2)= 0. Again, you must solve x-1= -4, x- 1= 0, and x-1= 2. Because -0= 0, the "-" before f is irrelevant.
Yes, saying -4+ 1= 3, -1+1= 0, 2+1= 3 is the same as saying 3-x = -4, -1-0= -1 and 3-1= 2 so, in a backward way, you are solving x-1= -4, x-1= 0, and x- 1= 2.

Quote:

(-3) x -1 = 3
0 x -1 = 0
3 x -1 = -3

I don't understand why you are multiplying by -1. The "-" outside f is multiplying a "y" value (y= f(x)) and has nothing to do with x values. As I said before -0= 0.

Quote:

#3 The maximum value of y = f(x) is 6. Find the maximum value of y = (1/3)f((1/2)x).[/B]
My work: (1/3)f((1/2)(6)) = 1. Not right though.

You replaced "x" by 6! But 6 is a value of "y", not x!
No matter what x is, the f value cannot be larger than 6 so (1/3)f(anything) cannot be larger than (1/3)(6)= 2.