[MOVED] graph f(x) = x^2 - 4x, g(x) = 1/f(x)

[kathy's]

This is the exercise I'm working on:

a. Draw the graphs of f(x) = x^2 - 4x and g(x) = 1/f(x) on the same grid.

b. Explain how the graph of f(x) = x^2 - 4x can be used to obtain the graph of y= g(x).

I have done part (a); it's part (b) that has me confused. All I know for sure is that the graph of f(x) has two zeroes, at x = 0 and x = 4. But I am unsure how to use this (or the rest of the graph) to answer the question.

Thank you for your help! :wink:

 

[stapel]

At a guess, they want you to look at the y-values of the graph of f, and put them under "1" to find the corresponding y-values for the graph of g.

For instance, since f(2) = -4, then g(2) would be -1/4, and so on.

Note: Since f(x) gets really big really quickly on either end, and since 1/(really big) is (really small), you draw a few points (like g(5) = 1/5) to either side of the zeroes of f(x), and then trail g(x) along the x-axis, close to y = 0. Also, since f(0) = f(4) = 0, and since you can't divide by zero, you'll have vertical asymptotes on the graph of g for x = 0 and x = 4.

Eliz.

 

[kathy's]

?

I don't need to draw a graph right but i need to write down the answer I am not sure what to write down as my answer cuz i can't really put it into words!

 

[stapel]

I am not sure what to write down as my answer cuz i can't really put it into words!

I can't help you write your paper, and if the explanation I gave didn't help, then you must be approaching this much differently (which is fine).

You might want to try explaining the process you're using to somebody else, and then write down what you eventually ended up saying.

Good luck!

Eliz.

 

[soroban]

Re: don't know were to post it!

Hello, kathy's!

a. Draw the graph of \(\displaystyle \,f(x) \:= \:x^2\,-\,4x\) and \(\displaystyle g(x)\:=\:\frac{1}{f(x)}\) on the same grid.

b. Explain how the graph of \(\displaystyle f(x) \:=\:x^2\,-\,4x\) can be used to obtain the graph of \(\displaystyle y\:= \:g(x)\)

The graph of \(\displaystyle \,y\:=\:x^2\,-\,4x\) is an up-opening parabola with x-intercepts \(\displaystyle 0\) and \(\displaystyle 4.\)

                  |
              *   |                   *
                  |
               *  |                  *
                * |                 *
      - - - - - - * - - - - - - - * - - - 
                  |  *         *  4
                  |      * *
                  |

The graph of: \(\displaystyle \L\,g(x) \:=\:\frac{1}{f(x)} \:=\:\frac{1}{x(x\,-\,4)\)

. . has vertical asymptotes \(\displaystyle x\,=\,0\) and \(\displaystyle x\,=\,4\)

. . and a horizontal asymptote \(\displaystyle y\,=\,0\)

                  |
                 *|               :*
                  |               :
                * |               : *
               *  |               :  *
             *    |               :    *
         *        |               :         * 
      - - - - - - + - - - - - - - + - - - - - -
                  |      * *      4
                  |   *       *   :
                  | *           * :
                  |               : 
                  |*             *:
                  |

This could be graphed point-by-point from \(\displaystyle f(x)\).

The values of \(\displaystyle g(x)\) are the reciprocals of the values of \(\displaystyle f(x)\).


Examples:

Since \(\displaystyle f(6)\,=\,6^2\,-\,4\cdot6\:=\:12\), we have the point \(\displaystyle (6,\,12)\) on the \(\displaystyle f(x)\) graph.
. . This corresponds to: \(\displaystyle \,\left(6,\,\frac{1}{12}\right)\) on the \(\displaystyle g(x)\) graph.

Since \(\displaystyle f(2) \:=\:2^2 - 4\cdot2\:=\:-4\), we have the point \(\displaystyle (2,\,-4)\) on the \(\displaystyle f(x)\) graph.
. . This corresponds to: \(\displaystyle \,\left(2,\,-\frac{1}{4}\right)\) on the \(\displaystyle g(x)\) graph.

 

[kathy's]

thanks!

thanks soroban and :wink: