hello, how would I plot a graph y = 2/x for -2<x<2 ?
Plot the extrema
$\displaystyle f(-2) = -1$
$\displaystyle f(2) = 1$
$\displaystyle f(0) = unde fined$
If you know the shape of the graph of $\displaystyle y=\frac{1}{x}$ you'll find it's much the same. Plot the two defined points above and draw the standard graph through
yeah thats the graph shape i get, its just i need to plot the other equation on the same graph, so i can find the place where they cross over, and they dont cross anywhere. the other equation is y = e^x and the line ive got already for that looks like its right, so where am i going wrong, or what am i missing ?
(0,0) is not a point on either graph.
On the attached graph $\displaystyle y=e^x$ is in green and $\displaystyle y=\frac{2}{x}$ is in blue.
As you can see they intersect and it's close enough to Mr F's answer so his is highly likely to be correct (and not just because he's Mr F )
The problem did not say "use a graphing utility to produce the plot".
Whenever you want to sketch a curve you should start out by finding its x- and y-intercepts by setting y and x equal to 0 respectively. For $\displaystyle y=\frac{2}{x}$ you can't find any finite value, so you know the graph doesn't cut the x & y intercept.
Then you should determine its asymptotes, You can calculate the vertical asymptote by inspecting where the graph would be undefined (i.e., $\displaystyle y=\infty$), obviously in this case the denominator is zero at $\displaystyle x=0$ so you have a vertical asymptote there. In order to find the horizontal asymptote you need to compute the limits of your function as $\displaystyle x \rightarrow \pm \infty$, if either limit has a finite value L, then the line y=L is a horizontal asymptote. $\displaystyle \lim_{x \rightarrow \pm \infty} \frac{2}{x} = 0$, so your horizontal asymptote is at $\displaystyle y=0$.
Using the values of y' you can then check where the graph is increasing (+ve) or decreasing (-ve). Values of y'' tells you where the graph is concave up or concave down. You have enough information to plot the graph.
As far as $\displaystyle y=e^x$ is concerned, you know that $\displaystyle e^x$ is the inverse of $\displaystyle ln x$, so its graph can be obtained from that of $\displaystyle ln x$ by reflection in the line $\displaystyle y=x$. Therefore its y-int is y=1 (as x=1 was the x-int of ln x), and the range of $\displaystyle e^x $will be the set of all positive real numbers.