[hydepark60]

Describe the transformations that have taken place and identify the new locations of the asymptotes.

[jcaron2]

Hydepark, when you write out a mathematical expression that has a fraction in it, make sure you put the denominator in parentheses. The order of operations in math dictates that you do multiplication and division before addition and subtraction. Therefore, what you wrote in your question literally means



However, I think it's clear that you meant



In the future, you should write that as f(x)= -(4/(x-2))+4.

Anyway, with that out of the way let's review what effects various transformations have on a plot:

First, when you replace all the x's in a function with (x-a)'s, that has the effect of shifting the plot a units to the right. Thus, you know that the plot of f(x)=1/(x-2) is the same as f(x)=1/x, except it's shifted to the right 2 units.

Second, when you multiply a function by a constant b, it has the effect of making the plot b times steeper (every y-coordinate in the plot gets b times bigger, kind of like if you drew the plot on a piece of rubber and then you stretched it out to be b times as tall along the y-direction). So now you know the plot of f(x)=4/(x-2) is 4 times as steep as the plot of f(x)=1/(x-2).

Third, a minus sign in front of a function has the effect of flipping the plot upside-down. So you know that the plot of f(x)= -(4/(x-2)) is just an upside-down copy of f(x)=4/(x-2).

Finally, whenever you simply add a constant c to a function, it has the effect of shifting the plot up by c units. Thus, the plot of f(x) = -(4/(x-2))+4 is the same as f(x) = -(4/(x-2)), but shifted up 4 units.

Anything that applies to the plot as a whole also applies to the asymptotes. So, since f(x)=1/x has an asymptote at x=0, the -2 in the denominator has the effect of shifting that asymptote over by 2 units to the right (along with the rest of the plot). The rest of the transformations only effect the y-coordinates of the data, so they have no impact on the vertical asymptote. (Since that asymptote is a vertical line extending from y=- to y=, stretching it, flipping it upside down, or shifting it up or down have no effect). Meanwhile, the horizontal asymptote at y=0 is only affected in this case by the last step where +4 was added to the function. That has the effect of moving it up 4 units.

[Unknown008]

Have you tried sketching the second function?

[Unknown008]

Sorry jcaron, didn't see your post before posting mine :o

EDIT: Removed the wrong piece out from my post =S

[jcaron2]

And something more to add to jcaron's post (which I didn't see until I posted my answer above) , the graph is steeper for values of y not between -1 and 1 (prior to the y-transformation) and less steep compared to the graph y = 1/x for the interval between -1 and 1.

:confused:

The slope is 4 times as large, regardless of y-value. Are you thinking of the case where you raise the function to an exponent, as opposed to merely scaling it? Or am I misinterpreting what you mean?

[Unknown008]

Oops, it's when you square it that the slope decreases, brain fart. :o

[jcaron2]

I know the feeling! :)