Pre-Calc. Question: Graphing Rational Functions

[bballwaterboy]

When you have a rational function, such as:

3x-5/x-1

After attaining things like the x and y intercepts and asymptotes, how do you know how many "pieces" of the graph there are? With linear functions/equations, you know it's a single line. Even quadratic graphs are a single piece - albeit a parabola. But with these rational type of functions/equations, there can sometimes be multiple pieces to graph.

How do you know how many there are?

 

[jedishrfu]

Most people would use calculus to determine where the slope is zero to give you an idea of where the function changes direction.

 

[Mark44]

When you have a rational function, such as:

3x-5/x-1

After attaining things like the x and y intercepts and asymptotes, how do you know how many "pieces" of the graph there are? With linear functions/equations, you know it's a single line. Even quadratic graphs are a single piece - albeit a parabola. But with these rational type of functions/equations, there can sometimes be multiple pieces to graph.

How do you know how many there are?

I presume you mean this:
$$f(x) = \frac{3x - 5}{x - 1}$$
What you wrote is $3x - \frac{5}{x} - 1$

Without using LaTeX, you should write the fraction using parentheses, like so: (3x - 5)/(x - 1).

There's a lot you can do without invoking calculus. Since the function is defined for all real x except x = 1, the graph will be in two parts, with the line x = 1 being the vertical asymptote.

By polynomial long division it turns out that (3x - 5)/(x - 1) = 3 + -2/(x - 1), so the graph is similar to the graph of y = -1/x with a vertical shift upwards and a horizontal shift to the right.

In addition to the vertical asymptote, there is also a horizontal asymptote, which is more evident in the 3 + -2/(x - 1) form I wrote.

About the only properties of this function that require calculus are the locations of the relative maximum and minimum points. Most everything else can be determined without the use of calculus.

 

[jbriggs444]

Can you see that a rational function will be continuous everywhere that it is defined? Intuitively, "continuous" means that its graph has no break.

Then the only question is how many places the rational function could be undefined.

 

[bballwaterboy]

I presume you mean this:
$$f(x) = \frac{3x - 5}{x - 1}$$
What you wrote is $3x - \frac{5}{x} - 1$

Without using LaTeX, you should write the fraction using parentheses, like so: (3x - 5)/(x - 1).

There's a lot you can do without invoking calculus. Since the function is defined for all real x except x = 1, the graph will be in two parts, with the line x = 1 being the vertical asymptote.

By polynomial long division it turns out that (3x - 5)/(x - 1) = 3 + -2/(x - 1), so the graph is similar to the graph of y = -1/x with a vertical shift upwards and a horizontal shift to the right.

In addition to the vertical asymptote, there is also a horizontal asymptote, which is more evident in the 3 + -2/(x - 1) form I wrote.

About the only properties of this function that require calculus are the locations of the relative maximum and minimum points. Most everything else can be determined without the use of calculus.

Hi, Mark44

Yes. I did mean f(x) = ... Sorry, I forgot to add the function notation before the equation. I'll definitely use your recommendation for how to write fractions in the future as well. I just didn't know how to write them with the line in this forum.

We don't do any sort of polynomial long division in my class. We get the vertical and horizontal asymptotes another way. But, assuming you have the asymptotes and the x & y intercepts, my question is still how do you graph this rational functions? I know there will be points on the intercepts, but then, how do you know:

a.) how many pieces there are
b.) where those pieces of the graph are located

?

Hope that questions makes more sense now. So, in other words, I've gotten all the relevant info. and I just literally want to know what to do with that info. in terms of graphing the rational function. Thanks very much!

 

[Mark44]

how do you know:

a.) how many pieces there are

The numbers that make the denominator zero determine how many pieces there are in the graph. In your function, there is only one value where the function is undefined. Therefore, the graph is in two pieces - one piece to the left of the discontinuity, and one piece to the right. This is what jbriggs444 alluded to in his post.

b.) where those pieces of the graph are located

See above.

 

[bballwaterboy]

The numbers that make the denominator zero determine how many pieces there are in the graph. In your function, there is only one value where the function is undefined. Therefore, the graph is in two pieces - one piece to the left of the discontinuity, and one piece to the right. This is what jbriggs444 alluded to in his post.
See above.

Ahh! Got it now! Thanks jbriggs444 & Mark44. This makes sense finally!

 

[AMenendez]

As Mark44 pointed out...
Find all of the vertical asymptotes (set the denominator equal to zero and solve for $x$--this will tell you the $x$-values through which the vertical asymptotes go).
Once you do this, you have a rough idea of where the graph of the function is "split" into pieces.

 

[Mark44]

As Mark44 pointed out...
Find all of the vertical asymptotes (set the denominator equal to zero and solve for $x$--this will tell you the $x$-values through which the vertical asymptotes go).

True enough for the example of this thread, but not necessarily true in general. For example the graph of this equation --
$$y =\frac{x^2 + 3x - 4}{(x + 4)(x + 1)} $$
-- has only one vertical asymptote. At the other discontinuity there's a "hole."

Once you do this, you have a rough idea of where the graph of the function is "split" into pieces.

 

[AMenendez]

Fair enough. Thanks for pointing that out.

 

[bballwaterboy]

True enough for the example of this thread, but not necessarily true in general. For example the graph of this equation --
$$y =\frac{x^2 + 3x - 4}{(x + 4)(x + 1)} $$
-- has only one vertical asymptote. At the other discontinuity there's a "hole."

Why is there only one VA? Would x = -4 and x = -1 both not be VAs? Thanks.

 

[jbriggs444]

If you factor the numerator, you will find that $x^2 + 3x - 4$ is equal to $(x+4)(x-1)$

 

[Mark44]

To elaborate on what jbriggs444 said, there's a "hole" discontinuity at x = -4. The factors of x + 4 in the numerator and denominator can be simplified to 1, as long as x is not equal to -4.