drawing graphs from equations?

Question

hi
i gotta draw a graph by a given equation so i fint the roots for the x values and i can do that with equations like y=(x-3)(x+5) cos then my values are just x=3 and x = -5 but then i cam across 2 questions i couldnt do
one is y= 12/x
and the other is
y= 3^x

please could someone explain how to put this into a graph
thnx xxx

Answer 1

just put numbers in for x and see what you get for y... For example

y = 12/x

x y
1 12
2 6
3 4
4 3


y=3^x


x y
0 1
1 3
2 9
3 27

Answer 2

This shouldn't be any more difficult than any other graphing question.

All you need to do is pick a value of x, and put it through the function to get the y value. Then you have the x,y coordinates for a point. You can keep doing this for the entire range you need to worry about.

So, for y = 12/x you could do this:

if x = 1, y = 12
if x = 2, y = 6
if x = 3, y = 4
if x = 4, y = 3
if x = 5, y = 2.4
if x = 6, y = 2

You'd also want to do this for the negative values of x (if you are supposed to) so:

if x = -1, y = -12
if x = -2, y = -6

and so on. .

You will also notice that 12 / 0 is infinite... So you will have a very high numbers as you approach 0 and no value as you approach 0.

For the other one y = 3^x just do the same thing and plot out some of the different coordinates.

Answer 3

y = 12/x is a hyperbola. Try graphing y = 1/x on a calculator to get the feel of what this kind of general function looks like. There can't be any roots to this equation, because if 0 = 12/x, then 0x = 12, which is impossible. However, as x approaches infinity or negative infinity, the y value approaches 0 (i.e. there is a horizontal asymptote at y = 0). As x approaches 0 from the right, y approaches infinity, and as x approaches 0 from the left, y approaches negative infinity (vertical asymptote at x = 0).

The other equation is an exponential function. This also has no roots, but as x approaches negative infinity, y approaches 0. It's very helpful again to look at the graph on a calculator or at least know what it should look like before trying to graph it. It starts off pretty level around 0 for negative x-values and then it increases substantially eventually.

Answer 4

y= 12/x

You need to know what the graph of y= 1/x looks like. You then just plot a few points, like plug in 1 for x, you get the point (1,12). You know from the 1/x graph what the graph looks like, you just know that the 12/x graph increases faster. Remember the asymptotes for this graph, the graph will never cross any of the axis, it has asymptotes at both x=0 and y=0.


y= 3^x

For this one, knowing what the e^x graph looks like is helpful. Especially since e is equal to about 2.718, so the two graphs will look very similar. Anyway, if you know what the e^x graph looks like, you know how the graph should turn out. For this specific grpah, again just find a few points that are really important. I suggest the y-intercept, since it is the point (1,3). Then just remember that as x approaches negative infinity the graph gets closer to the x-axis (or zero) and the as x approaches positive infinity the graph just keeps going up.

Answer 5

Hey,

Sometimes when I need to get an idea of the behavior of an equation I graph it in Microsoft Excel. In the column A enter the indepedent variable values. -3, -2, -1, 0, 1, 2, 3 etc. Then in cell B1 enter the equation ie "=12/A1" or "=(A1-3)*(A1+5). The cell B1 will use the number in the cell A1 with the equation you enter to determine the dependent variable values. Next copy cell B1 (containing your formula) to the rest of column B. Graph the two columns and you got it!! I can send you a sample spreadsheet if you like. I don't how familiar you are with Excel.

Answer 6

Plot the 2 graphs. Then, find the point where the two graphs meet. Algebraically, the answer should be: 4x -3 = 6 - 2x 4x -3 + 2x = 6 - 2x + 2x 6x - 3 = 6 6x - 3 + 3 = 6 + 3 6x = 9 x = 9/6 = 3/2 = 1.5

Answer 7

Just pick random numbers for x and calculate out what y would equal and plot them on your graph. For y=12/x is easy. Anything divided by 0=0. y=3^x never hits 0 so you have to just again pick numbers and put it in and get your answer and just plot it on the graph. You should probably also look into getting a graphing calculator.

Answer 8

y = 12/x is not defined at zero. It has a positive asymptote on the positive side of zero and a negative asymptote on the negative side of zero. It looks like a hyperbola.

y = 3^x never goes below zero. It starts near zero at negative infinity, and crosses the y-axis at (0,1), and then shoots off to positive infinity to the positive infinity side.