how do u graph a function without using a graphing calculator in advanced calc?

Question

the equation is ---> 1/(x-1)(x-3) that is 1 divided by the polynomial (x-1)(x-3) .....if it is possible can u give me the step by step instructions ****w/o using a graphing calculators please

Answer 1

The answers that suggest plotting points should be disregarded - for advanced calc you need to be a lot more sophisticated. This is a rational function. The person who suggesting "inverting" the parabola is conceptually correct, but that's hard for people to visualize, especially with points where the denominator is not defined.

Start with asking the domain of the function- it's defined everywhere except x = 1 and x = 3, and near those two points there will be an "explosion". The function is never zero, so it never crosses the x-axis. So it will have three chunks. For huge x it will look like 1/(x^2) > 0 and going
to zero as z goes to +- infinity. Then find the critical points (there has to be one in the interval (1.3) ) plot each critical point, and then you should be able to graph - and if you like plot a few points for security sake.

Answer 2

There are already quite a few good answers. The inversion as an reflection into the line y=x is incorrect. It works for inverse function. However, (x-1)(x-3) and 1/(x-1)(x-3) are not inverse of each other.
However, this is close to an easy way to get the graph, i.e. by finding the reciprocal.

(x-1)(x-3) is a parabola, crossing x axis at (1,0) and (3,0), minimal point at (2, -1).

Now, consider the reciprocals of the y values of the parabola:
When y >= 1, 0<1/y<=1, and vice-versa,
when y <= -1, -1<=1/y<0, and vice-versa,
1/y never a zero, but when y approaches infinity, 1/y approaches zero.
When y = 0, 1/y is undefined, but when y is close to 0 but negative, 1/y approaches negative infinity, and when y is close to 0 but positive, 1/y approaches positive infinity.

So, jot down the relevant points, sketch something to indicate approaching events to zero, + or - infinity, and then join them up smoothly (since the 1st derivative is defined except for the disjoint places at x = 1 and x = 3).

Thus, the graph has 3 portions, for x<1, 13.

Answer 3

Just calculate what y will be if x=0, 1, 2, 3, 4,.... etc. and plot the dots in a graph.

For example, when x=0, y=1/[(0-1)(0-3)] = 1/3. The first dot on the graph is (0,1/3) and so on. There are no solution when x=1 and x=3 because the denominator equals 0.

Answer 4

First solve the equation for when y = 0. That will tell you then it crosses the x axis.

Then solve the equation fofr when x = 0. That will tell you when it crosses the y axis.

Then take the first derivative of the function, and find where y=0 for that function. That tells you where the highs and lows are.

Also, find where the denominator in any of the function = 0. That will tell you the vertical asymptotes. And where there are any square roots go negative is another asymptote. There might be other times where the function is undefined, too. Sometimes you can have functions asymptotic to other lines or curves, too.

I know this sounds complicated, but it really isn't if you break the function down.

Answer 5

I'm assuming that the fraction there is all equal to 'y' (y = 1/{[x-1][x-3]}), in which case my answer would be similar to Gabe's (the first response I saw):

draw a cross (like a plus-sign, but with the vertical-line longer at bottom); to the left of the vertical center, write an 'x' above the horizontal line and the numbers '-2,' '-1,' '0,' '1,' and '2' below the horizontal line; then plug in '-2' for 'x,' solve for 'y,' and put that answer under the 'y' on the table next to 'x:-2;' do likewise for all the following 'x's.

When you have the five 'y's corresponding to the five 'x's, map them out on the chart.

Actually, that particular equation has at least two sets of answers: x^2 - 4x + 3 ([x -1][x - 3]) must be greater-than zero, meaning x^2 - 4x must be greater-than -3 ... after that, I'm not sure ... the point is, you'll have a negative set and a positive set. Pick 'x'-numbers at random, and plug ` em in `til it works to find both sets.

Answer 6

You need to do a lot of excercise.

1. First step is to determine the shape of the graph. If linea equation,
y = mx + c
the graph is linear cut at the point c when x = 0
the graph is having an upward trend if m is positive and vice versa

2. If the graph is curve, then it shall be y = ax^2 + bx + c.
logically once you see ^2 and above, the graph is curv type.
if a is negative then the graph is a smilling face and vice versa
if x is 0, then y will cut at c if x = 0
using this formula x = {-b^2 + root(4ac)}/2a and {-b^2 - root(4ac)}/2a to find value of x if y =0

if 4ac is a negative figure, then the graph is not intersecting with x axis.

Hope this help

Answer 7

Calculate the values of the equation for various values of x. For example, figure out what your function is equal to when x=1,2,3,4, etc. You use the set of values for X that are most relevent; it's the equivilant of setting the range of X values in a graphing calculator.

Each of these represents a set of coordinates, with the X coordinate being equal to the value of X that you used, and the Y coordinate being the value of the function using that value of X.

Then, plot these points on a graph.

Then, connect the dots.

Thus, you have a graph. That's basically how a graphing calculator does it, too.

You will, of course, have to understand the function you're graphing, so that you know how to recognize asymptotes and such (and thus choose the proper values of X to use, so that you connect the dots properly, and get a proper result for the graph).

Answer 8

if you can graph (x-1)(x-3), which is a parabola, than you can graph 1 divided by the same. Anything with a one over it is an inverse of the stuff on the bottom, so your graph of (x-1)(x-3) would be reflected in the line y=x. That is the easiest way that I can tell you to graph it. Just graph the bottom, and then switch all the x and y values.

Answer 9

When graphing functions by hand, watch for significant events.

Since division by zero is undefined, You function will be undefined at X= 1 and x=3.

Other than that you can pick values of X and make yourself a table from which to draw the graph.

Answer 10

First you need to multiply out the bottom number...It should be x^2 * -4x * -3. Then you just plug in a few points on the graph to get a general idea of how the function should look.