Functions Help (for test tmr.)

mikexz

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Feb 21, 2006
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How do I solve (graph) functions which are a combination of 2 basic functions. More specifcally:

f(x)= 1/x^2?

Basically I am asking, how do I know which function I should be making the transformations to. (example. x^2-2 In this one I know that I would be making the transformations to a parabola.

I am also having trouble with this one:

f(x) = 1/x^2-x-6

-I think that it looks like a parabola but I am not sure because I don't know how the -x affects it.

Last problem. For reference-http://en.wikipedia.org/wiki/Functional_composition

g o f = h

f(x)=x+4 & h(x)= 3x^2+3x+2


I know how to do questions in which the known function is first but not when it's second.

Thanks for your help
 
f(x)= 1/x^2?

Basically I am asking, how do I know which function I should be making the transformations to. (example. x^2-2 In this one I know that I would be making the transformations to a parabola.

I am also having trouble with this one:

f(x) = 1/x^2-x-6

-I think that it looks like a parabola but I am not sure because I don't know how the -x affects it.

Hello Mikexz,

Neither of these functions are parabolas, nor do they look like parabolas.

Neither f(x)= 1/x^2, nor f(x) = 1/x^2-x-6 have simple “parent” functions that you just translate in some direction (if that’s what you were asking/implying), except you could consider f(x)= 1/x^2 to be a “parent” function itself. To graph these functions, it is required that you identify vertical and horizontal asymptotes.

For the first function, f(x)= 1/x^2, you have the x and y axes as asymptotes. (Note that x cannot equal 0 in this problem, so you have a vertical asymptote at x = 0.) Plug in some trial numbers for x and find the corresponding y values and plot them to see what the curve looks like.

For the second function, f(x) = 1/x^2-x-6, I’m guessing you actually meant f(x) = 1/(x^2-x-6). If so, the denominator can be factored to (x-3)(x+2). Set each factor equal to zero and solve for x to find values that x cannot have in this problem (meaning not in the domain of x). Obviously x cannot equal 3 or –2. Therefore you have vertical asymptotes at these values of x. Again, plug in some values for x in the three different regions of the graph defined by these asymptotes to find the shape of the curve in each region.

There is a little more to it (evaluating asymptotes) than what I’ve given you here, but this is the “short and sweet” explanation. Hope this helps.
 
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