Let's take a look at rational functions. We'll start with the definition.
A rational function is a function of the form: f of x equals g of x over h of x,
where g and h are polynomial functions such that the one in the denominator is not equal to 0.
There's a special case of rational functions where h of x equals c,
c being a non-zero real number. We'll consider those just to be polynomial functions
and not treat them as rational functions.
Two important features of every graph of every function we've ever studied
are domain and intercepts. The domain for a rational function
is all real numbers except for any which make the denominator, h of x, equal to 0.
The intercepts are found just as we always find intercepts.
To find the y-intercept, if it exists, let x equal 0 and evaluate the function.
If there are any x-intercepts, you find them by solving f of x equals 0.
We should note that rational functions may or may not have a y-intercept.
If there is one, there is only one.
x-intercepts may or may not exist as well,
and a rational function may have zero, one, or many x-intercepts.
In this example, we're given a rational function,
and we're asked to find its domain and any intercepts.
Recall that the domain of a rational function is all real numbers
except any that make the denominator equal to 0.
Let's start by finding any real numbers that make our denominator 0.
Solving this, we see that x equals 15 and x equals negative 15
are the two values that make our denominator equal to 0.
We'll need to exclude these from our domain.
We're usually asked to write our domains in interval notation, so let's do that.
We want the set of all real numbers less than negative 15,
between negative 15 and positive 15, as well as those that are greater than 15
Here's the domain of our function.
Now let's talk about intercepts. First, we're going to find the y-intercept.
To do this, we let x equal 0, so we need to evaluate f of 0.
That's going to give me 0 plus 1 over 0 squared minus 225.
Simplifying, we get negative 1 over 225. That is our y-intercept.
Now, the x-intercepts: if there are any, we find them by letting y equal 0.
y is f of x, so we need to solve 0 equals x plus 1 over x squared minus 225.
The first step in solving this is we want to multiply both sides
by the denominator, x squared minus 225. So I end up with 0 equals x plus 1.
x equals negative 1 is our only x-intercept.
Notice that you might see a shortcut you might want to take.
Since the first thing we did was eliminate the denominator,
it turns out that the x-intercepts are always the solutions of 0 equal to your numerator
as long as there aren't any common factors.
Let's look at another example. Again, we're given a rational function,
and we're asked to find its domain and any intercepts.
To find the domain, we start by finding any values
that will make our denominator equal to 0 so that we can exclude them.
We factor out the common factor of x.
x squared plus 4 can't be factored any further over the set of real numbers,
so the only number that needs to be excluded from our domain is x equals 0.
Our domain in interval notation is all real numbers less than 0
along with all real numbers greater than 0.
Now let's look at the intercepts. We've just excluded x equals 0 from our domain,
so when I go to find the y-intercept, there can't be one since x can't be equal to 0,
but there may still be some x-intercepts. Let's look at our function.
I want to remove any common factors before I look for my x-intercepts,
so both the numerator and denominator have a common factor of x.
Dividing those out, I'm left with x plus 5 over x squared plus 4.
The x-intercepts will be any values of x that make our numerator equal to 0,
so x equals negative 5 is our x-intercept.
Many rational functions have vertical asymptotes.
Let's talk about what a vertical asymptote is.
I've drawn you several examples. The vertical line x equals a
is a vertical asymptote of a rational function or any function
if at least one of the following occurs. In this case,
the function values, or the y-values, are getting larger and larger as you get close to a
from the right-hand side. We say that the function values are approaching infinity
as x approaches a from the right.
In this case, our function values are getting smaller and smaller as we approach a
from the right again. We say that our function values are approaching negative infinity
as x approaches a from the right.
In this case, we're approaching a from the left,
and again our function values here are rising,
so we say that the function values are approaching infinity as x approaches a from the left.
Finally, in this picture, our function values are getting smaller and smaller
as we approach a again from the left,
and we say that f of x approaches negative infinity as x approaches a from the left.
If any one or more of these occur at x equals a,
then that is a vertical asymptote for the graph.
A rational function in particular will have, as long as g and h have no common factors,
will have a vertical asymptote at x equals a whenever h of a equals 0.
Let's talk more about vertical asymptotes.
Here we're given a rational function, and we're asked to find all of its vertical asymptotes
and then create a rough sketch
to describe what the graph looks like near each of the asymptotes we find.
Recall that a rational function will have a vertical asymptote at any value a
that makes the denominator 0 as long as it doesn't also make the numerator 0,
so we need to remove common factors before we begin.
In the numerator, I can factor out the common factor x.
The denominator's a trinomial that can be factored.
We're going to use factors of 3, which are 1 and 3,
and we need that 3 to be negative to make this work out.
Okay, so we do have some common factors.
Dividing out the common factor, we're left with x over x plus 1.
The only number that makes our denominator equal to 0 is negative 1,
so our only vertical asymptote is going to be x equals negative 1.
Alright, now you probably are thinking, but what about the 3?
3 is not in the domain because it makes our original function undefined.
Well, that's a special case that will come up a little bit later in our video.
We've just determined that this rational function has one vertical asymptote
at x equals negative 1. Let's sketch that.
Our next task is to determine how does the function behave
as we get very close to this asymptote on either side. We will use test values.
Start by picking numbers just a little bit to the left of negative 1.
You want to be close, so don't pick negative 2 or something smaller than that.
Let's start with say negative 1.3.
Substitute negative 1.3 in place of the x in your reduced rational function.
This is the first step. Simplify the denominator,
so f of negative 1.3 is 13/3. That's a little bit more than 4.
We need to go a little bit closer to try to get a better handle on the behavior,
so now choose x to be negative 1.2. Substitute into our reduced rational function.
Your first step will be negative 1.2 over negative 1.2 plus 1.
Subtract in the denominator, and this value is 6. f of negative 1.2 is 6.
Getting a little bit closer, let x be negative 1.1.
Again, substitute into the reduced rational function.
Working that out as we did before, this one is 11,
so now we look to see is there a pattern?
This is a litter bit more than 4. This one's bigger at 6. This one's bigger at 11.
If we're not sure of the pattern, we could pick more test values closer and closer.
This certainly seems reasonable that our values are getting larger and larger
as our x gets closer and closer to negative 1.
Our function, our y-values, are getting bigger as we get closer to negative 1 from the lefthand side.
Now what about from the right? Again, we want to choose values close to negative 1,
so we can't start way over here at 0 or 1. We need to be much closer in.
We'll start with say negative 0.7. Let x be negative 0.7.
Substitute that for x in our reduced rational function.
That's going to be negative 0.7 over negative 0.7 plus 1.
Simplify the denominator. You get negative 7/3.
Repeat that for negative 0.8,
substituting again in our reduced rational function and simplifying to get negative 4.
Then finally, we'll use negative 0.9. Substitute in the reduced rational function and simplify.
Now let's look for a pattern coming from the right side of x equals negative 1.
This is negative 3 and 1/3, negative 4, negative 9,
so it seems reasonable that our values are getting smaller and smaller
as x gets closer and closer to negative 1 from the right.
Okay, so these are our y-values, our function values, getting smaller and smaller
as x approaches negative 1 from the right.
This graph describes the behavior of our function f very close to the asymptote,
and I'd like to point out that if you're not convinced at this point,
you can always pick more and more values.
The correct number to choose is what shows you the pattern.
Now let's take a look at horizontal asymptotes. Here's the definition:
If the values of f of x approach a fixed real number H
as x approaches infinity or as x approaches negative infinity,
then y equals h is a horizontal asymptote of f of x.
Make sure you understand x approaches infinity
means that your x-values are getting larger and larger and larger,
so we will look towards the right side of our graph.
As x gets larger and larger and larger,
notice that these function values are getting closer and closer to the line y equals H.
As x approaches negative infinity
means that our x-values are getting smaller and smaller and smaller,
so we're looking towards the left side of our graph.
As the x-values approach negative infinity, notice that these y-values
are also getting closer and closer to the horizontal line y equals H.
Here's a second example.
As the x-values get larger and larger as they approach infinity,
the function values are getting closer and closer to H.
The graph is getting closer and closer to the line y equals H,
so we have another horizontal asymptote.
As x approaches negative infinity, again the function values are getting closer and closer to H,
and the graph itself is getting closer and closer to the line y equals H.
The graph on the left is a little bit different. It still has a horizontal asymptote,
but notice that you only need one side to look like an asymptote. Right?
The y-values get closer and closer to y equals H as x approaches infinity.
That's enough to make this a horizontal asymptote.
Notice that the graph actually crosses over that horizontal asymptote over here somewhere.
That's okay. A function may or may not cross its horizontal asymptote.
It might cross it many times. That's okay.
A function never crosses its vertical asymptotes.
Let's talk more about horizontal asymptotes.
Here's a rational function of the form g of x over h of x where g and h are polynomials,
and we'll write them out into this form where we can see all of the terms of the polynomials.
The a sub n, a sub of n minus 1, and so on are just any real numbers.
n is the degree of the polynomial g in the numerator,
and m is the degree of the polynomial h in the denominator.
A rational function can have many vertical asymptotes,
but a rational function can have at most one horizontal asymptote.
Whether or not a horizontal asymptote exists, and where it is, if it does exist,
depends on the relationship between m and n.
If m is greater than n, then y equals 0 is the horizontal asymptote for the function.
If m equals n, then y equals the ratio of the leading coefficients will be your horizontal asymptote.
Finally, if m is less than n, then the function has no horizontal asymptote.
Next, we'll look at some specific examples.
In this example, we'll consider three rational functions
and look for their horizontal asymptote if it exists.
Remember that the relationship between
the power of the numerator and the power of the demoninator, n and m,
will help us to determine if there is a horizontal asymptote and where it is.
In this first one, the degree of the numerator is 3. The degree of the denominator is 1,
so since m is less than n, there is no horizontal asymptote.
In our second example, the degree of the numerator is 1. The degree of the denominator is 3,
so m is greater than n. The horizontal asymptote will be y equals 0.
Finally, let's look at this one. Please don't be tricked.
The degree of the numerator is 2 and so is the degree of the denominator.
Since m equals n, the horizontal asymptote is y equals the ratio of the leading coefficients,
so that would be positive 2 over negative 17.
Let's look at two special rational functions, their graphs, and some of their properties.
The first function is f of x equals 1 over x. 0 makes this denominator equal to 0,
so it has to be excluded from the domain, so our domain here is all real numbers less than 0
along with all real numbers greater than 0 written in interval notation.
The range is the same thing: y can't equal 0.
It has no intercepts, and it has two asymptotes.
The vertical asymptote is at x equals 0. The x equals 0 is the y-axis.
It has another asymptote: the horizontal asymptote at y equals 0. y equals 0 is the x-axis.
You notice when you look at this that it is symmetric about the origin,
so this is an odd function. Because it's an odd function,
it's true that f of negative x is equal to negative f of x.
Now let's look at f of x equals 1 over x squared.
Again, x cannot equal 0, so we exclude 0 from the domain.
Now look at the range. x can be any positive real number, and that's it,
so the range will be 0 to infinity with a parenthesis on the 0 since it's not included in the range.
The graph has no intercepts. It also has a vertical asymptote: x equals 0, the y-axis.
It has a horizontal asymptote: y equals 0, which is the x-axis.
This graph has a different kind of symmetry. It's symmetric with respect to the y-axis.
This is an even function, and for even functions f of negative x is the same as f of x.
We will use these two functions and transformations to graph some other rational functions.
In this example, we're asked to use those two special rational functions
to graph rational functions using transformations.
Let's look at the first one: f of x equals negative 1 over x plus 4.
What type of transformations will you need to do?
Correct. The negative in front of 1 over x tells me that there's a reflection across the x-axis,
and the plus 4 sort of outside tells me the graph should be shifted up 4 units.
We'll start with the reflection across the x-axis.
This point originally is at (1, 1).
When we reflect that across the x-axis, the new y-coordinate will be negative 1.
The x doesn't change.
The behavior of the graph is still that it has asymptotes at the x-axis and the y-axis.
This point, the y-coordinate right now is negative 1.
When we reflect it across the x-axis, the new coordinates will be (-1, 1).
Note that reflecting the graph across the x-axis doesn't affect the asymptotes at all.
Okay, now we'll take our reflected graph and shift everything up 4 units.
As soon as we start moving up or down,
then we're moving our horizontal asymptote that amount as well,
so instead of y equals 0, our asymptote will now be at y equals 4.
Now we can move the points up 4 units. Sketch the graph of that little section,
and then we have one more point to move up 4 units here.
Instead of the y-coordinate here at 1, we move that up 4 units. It'll now be at 5.
We draw that section of the graph using the asymptotes to guide us.
This is the graph of y equals negative 1 over x plus 4.
The vertical asymptote is still x equals 0.
The horizontal asymptote is y equals 4.
There are no y-intercepts. There is an x-intercept.
If we wanted to find its value, then we know techniques for doing that.
Now let's look at this function. This is 3 over x plus 2 squared.
What sort of transformations will we need to do?
The plus 2 inside sort of with the x
tells me that I need to shift the graph to the left 2 units,
and the 3 tells me the graph will be stretched vertically by a factor of 3.
We need to do the horizontal shift first.
If we're moving the graph left or right, that's going to affect our vertical asymptote.
In this case, we're going to move our vertical asymptote from 0 to the left 2 units,
so our vertical asymptote is x equals negative 2.
Then we can move the points and the pieces of our graph to the left 2.
We've changed our vertical asymptote with this transformation.
Notice that we have not changed the horizontal asymptote.
Alright, there's one last transformation.
Vertical stretching changes the y-coordinates of the points.
We can take the y-coordinates of this graph.
We're going to multiply them by 3 to obtain the y-coordinates of our new graph.
If we look at this point to start with, the y-coordinate is 1. 1 times 3 is 3.
We should go ahead and draw that vertical asymptote.
Stretching does not affect the vertical asymptote.
It also doesn't affect the horizontal asymptote,
so, when we draw this piece of the graph, we're still approaching the x-axis
from the top as x goes towards infinity.
Then over here, the y-coordinate is 1.
Multiply that by 3. The new y-coordinate will be 3.
Okay, so for this graph, to summarize:
our vertical asymptote is now at x equals negative 2.
The horizontal asymptote is still y equals 0.
It does have a y-intercept. It might not really be at 1.
Let x equal 0, and you solve for your y-intercept.
It doesn't have any x-intercepts.
Let's reconsider a function from earlier in this video:
f of x equals x squared minus 3x over x squared minus 2x minus 3.
The first thing I want to do is find its domain.
For a rational function, the domain will be all real numbers
except any that make the denominator equal to 0.
We'll start by factoring the denominator and looking for the numbers that make it 0.
We see after factoring that x equals 3 or x equals negative 1 will make this denominator 0.
Those are the values we need to exclude from the domain.
The domain will be all real numbers less than negative 1,
between negative 1 and positive 3, and greater than 3.
Now let's talk about vertical asymptotes.
A rational function has a vertical asymptote at x equals a
if a makes the denominator 0 but not the numerator.
We need to remove any common factors. Factor the numerator.
We see that there is a common factor of x minus 3.
Reducing to lowest terms, we're left with x over x plus 1.
We know that there's a vertical asymptote now at x equals negative 1,
but what about x equals 3?
It's not in the domain, but there is no vertical asymptote.
As it turns out, there's a hole in the graph at x equals 3.
We can find the coordinates of this hole using our reduced rational function.
When x is 3, we would get 3/4, so the hole is going to occur when x is 3 with a y-value of 3/4.
What we'd like to do next is see this on the graph and how it affects the graph as a whole.
Another piece of information we need to be able to draw the graph
will be the horizontal asymptote. We need to look at the reduced rational function,
and we see that the degree of the numerator and the degree of the denominator are both 1.
The horizontal asymptote will be the ratio of the leading coefficients.
In this case, that's y equals 1.
Now let's see what we can do with the graph.
The horizontal asymptote is y equals 1, so we'll start there.
The vertical asymptote we found is at x equals negative 1.
So far, the only other thing we know is there's a hole in the graph
at this point, so I'll go ahead and start with that.
This was 1, so 3/4 is about right here.
How am I going to finish the graph? At this point, I just need to use some test values.
Let's start on this side of the graph. Let's say we take x equal to 0.
That's 0 over 0 plus 1, so 0.
We know that as we get close to the vertical asymptote, the behavior should look like this.
If we are not sure, we can plot another point say where x is negative 1/2.
Then as we move this direction, we have that horizontal asymptote,
so we just leave out that hole.
Again, you can plot as many points as you need to
to be sure that you have the shape correct.
What about over here? Let's say this is negative 1,
so we'll take negative 2.
This is the point (-2, 2), and we can plot more points again if we need to
to be sure of the behavior.
We know that we have a vertical asymptote here going up
and then the horizontal asymptote here.
We have a graph of a rational function with a hole in it.
In calculus, this hole is called a removable discontinuity.
This means that it's one point that divides the graph into separate pieces,
but it can be filled in with just that one point
to make it continuous and smooth on that portion.
With sketching the graph of a rational function,
we want to use every tool at our disposal.
The first thing we want to find is its domain.
Factoring the denominator, I see that x equals negative 2 or x equals positive 2
both will make the denominator equal to 0,
so I'm going to exclude those values from the domain.
The domain will be all real numbers less than negative 2,
between negative 2 and positive 2, and greater than 2.
Next, I want to check for any common factors. There are none.
This tells me that I have no removable discontinuity.
Now we'll check for symmetry.
To check for symmetry, we want to find f of negative x.
Replacing the x with negative x, I have 1 over negative x squared minus 4.
Now be very careful with those parentheses. The negative x is what is squared.
Simplifying, this is the same as 1 over x squared minus 4.
We see the f of negative x is in fact the same thing as f of x.
This means that our function is even.
Now I want to find any intercepts.
To find the y-intercept, we let x equal 0,
so that means we want to find f of 0.
We have a y-intercept at negative 1/4.
To find the x-intercepts, we want to solve f of x equals 0.
Multiplying both sides by x squared minus 4, we get 1 equals 0,
but of course 1 doesn't equal 0. That means that there are no x-intercepts.
Now we need to find any asymptotes.
The function will have a vertical asymptote at any value that makes the denominator 0
but doesn't make the numerator 0, so at negative 2 and at positive 2.
The horizontal asymptote depends on
the degree of the numerator related to the degree of the denominator.
The degree of the denominator m is greater than the degree of the numerator, which is 0,
so the horizontal asymptote is y equals 0.
To draw the graph, we'll start with the things we've found so far.
The vertical asymptotes are x equals 2 and x equals negative 2.
The horizontal asymptote is y equals 0, which is already there. That's the x-axis.
Do I have any intercepts? I had one. I have a y-intercept at negative 1/4.
That's all I know so far, so I'll have to use some test values
to help me determine the shape of the graph.
Let's start on the far left side.
What is the graph doing when I'm close to the vertical asymptote on the left side?
We'll start with negative 3.
Substituting that into our function, we have 1 over negative 3 squared minus 4.
9 minus 4 is 5, so that's 1/5,
and we need to know if we need to plot more points.
The answer is: probably, but you can also reason a little bit,
knowing a little bit about the general shape of your rational functions.
We can't go across the x-axis because we don't have any more x-intercepts.
We know that there's supposed to be an asymptote,
so as x goes towards negative infinity, we're getting close to the asymptote.
Here, if we maybe plotted one more point in the middle,
we would see that these values are rising
and getting to the vertical asymptote from the left.
Here, we'll need to plot a couple of points. We'll go with negative 1.
1 minus 4. That would be negative 3, so this is negative 1/3.
Using the asymptote that I know exists here,
if I want to plot a few more points and be certain I have the right shape, that's fine.
Then you might be thinking: I've got more points to plot,
but, in reality, I have one other piece of information available.
This is an even function. That means that f of negative x is the same as f of x,
so I know that if f of negative 3 is 1/5, then f of 3 is also 1/5.
If f of negative 1 is negative 1/3, this is going to be the same for f of 1.
Again, I'm using the asymptotes to help. I can plot more points as I need to.
Okay, it's an even function. It should be symmetric with respect to the y-axis,
and it certainly looks like it is.
Let's graph another rational function.
This is the function f of x equals 3x over x squared minus 25.
As always, we'll start by finding the domain. Factor the denominator,
and the numbers that make our denominator equal to 0 will be negative 5 and positive 5.
These are the numbers to be excluded from the domain of the function.
Our domain is all real numbers less than negative 5,
between negative 5 and positive 5, and greater than 5.
Now look at the numerator. We can't factor, so we see that
we don't have any common factors between the numerator and the denominator.
That means we don't have to worry about any removable discontinuities.
Now let's talk about symmetry. We need to find f of negative x.
That's 3 times negative x over negative x squared minus 25.
The numerator is negative 3x. The denominator is the same as x squared minus 25.
We see that in this example, this is not the original function we had.
In fact, it's the opposite of the original function we had,
so f of negative x is the same as negative f of x.
This means that f is an odd function, and it's symmetric with respect to the origin.
Now let's look for any intercepts. To find the y-intercept, we let x equal 0.
We need to find f of 0.
Substituting 0 for each x and then simplifying, we get 0.
For the x-intercepts, we've already seen that
there are no common factors between the numerator and the denominator,
so the x-intercepts will be any values of x that make our numerator equal to 0.
We need to solve 0 equals 3x, our numerator,
so we see that x equals 0 is our x-intercept. Now let's talk about asymptotes.
Vertical asymptotes occur at any value of x that makes our denominator 0
that does not also make the numerator 0.
That's negative 5 and positive 5.
Our vertical asymptotes are x equals negative 5 and x equals 5.
Horizontal asymptote: whether it exists and what it is depends on
the relationship between the degrees of the numerator and the denominator.
The numerator is degree 1. The denominator is degree 2.
Since the degree of the denominator m is greater than the degree of the numerator,
then y equals 0 is your horizontal asymptote.
Let's put what we know so far on the graph.
We found vertical asymptotes: x equals 5 and negative 5.
We found the horizontal asymptote is y equals 0, which is the x-axis.
The x- and y-intercept are both 0.
This is all we have so far. We'll have to use test values
to determine the rest of the shape of the graph.
Let's start on the right-hand side. Choosing a number bigger than 5, perhaps 6.
We're going to substitute in our given function.
11 times 1 is 11, so when x is 6, y is 18/11.
That's close to 2 but a little bit less than 2.
Okay, using the asymptotes as a guide and plotting more points if you wish,
we will graph this section of the function.
Alright, we found earlier that our function is an odd function,
and that means that f of negative x is the same as negative f of x.
Using that, we can find f of negative 6. It has to be the opposite of f of 6.
We can plot that point, and again using the asymptotes and perhaps plotting more points,
we can finish the shape of our graph in that section.
Now what about in between the asymptotes?
I'll choose x equals 2. Substitute into the given function.
7 times negative 3 is negative 21, and reducing that is negative 2/7.
6 divided by 3 is 2.
21 divided by 3 is 7. Okay, so this is negative 2/7.
Then finally, using the fact that our function is odd one more time,
we know that f of negative 2 is negative f of 2, so we get positive 2/7.
Okay, we know that we have a vertical asymptote here.
We cross through our x-intercept, and we have another vertical asymptote.
Again, we can plot as many points as necessary
to be confident that that is the shape of our graph.
Let's look at another rational function: f of x equals 3x plus 1 over x squared plus x minus 20.
We start by finding the domain, so we need to factor the denominator.
5 times 4 is 20, so if we use plus 5 and minus 4 on the other one, this should work.
The numbers excluded from the domain will be negative 5 and positive 4.
Our domain is all real numbers less than negative 5,
between negative 5 and positive 4, and greater than 4.
There are no common factors in the numerator and denominator,
so we don't have any removable discontinuities.
We'll move on to finding symmetry.
To find the symmetry, if there is any, we find f of negative x.
Okay, so replacing x with negative x
and then cleaning it up a little bit, this is negative 3x plus 1 over x squared minus x minus 20.
What we notice is that this is not equal to the original function,
nor is it the opposite of the original function.
This graph has no symmetry. It's neither even nor odd.
Now let's talk about intercepts.
We'll go ahead and find the y-intercept first.
Let x equal 0, so we're finding f of 0: negative 1/20.
Now we'll find the x-intercept.
We saw that there were no common factors in the numerator and the denominator,
so that means we can find the x-intercepts by finding
the numbers that make our numerator equal to 0. Solve 3x plus 1 equals 0.
Our x-intercept is negative 1/3. Now let's find the asymptotes.
We have vertical asymptotes at x equals negative 5 and x equals 4.
To find the horizontal asymptote, we look at
the degree of the numerator and the degree of the denominator and compare them.
The degree of the denominator is 2. The degree of the numerator is 1.
Whenever the degree of the denominator m is greater than n,
then y equals 0 is our horizontal asymptote.
As you can see, I've gone ahead and sketched the vertical asymptotes:
x equals negative 5 and x equals 4. Our horizontal asymptote is y equals 0.
Let's plot our intercepts. The y-intercept is negative 1/20.
The x-intercept is negative 1/3, and that's all I have.
I need test values. I've picked four. 5 is in the far right-hand section.
Substituting 5 into our given function, we come up with the value of 8/5.
Again, using the asymptotes to guide us as well as plotting as many points as we'd like,
we can draw that section of the graph.
x equals 1 is in this section. Substituting that into the given function
and evaluating it, we come up with negative 2/9.
Let's go ahead and plot this point when x is negative 1.
By substituting in the given function, we find the y-value is 1/10.
Now, what does the graph look like in this portion?
Well, there's a vertical asymptote here,
so it seems reasonable that the graph is going to make this sort of shape.
If we're not convinced of that, we simply need to plot more points here or on this side.
The last test point we'll consider is x equals negative 6.
Substituting that into the given function and evaluating this, we come up with negative 17/10.
Again using the asymptotes to guide us as well as plotting more points as we wish,
here's that section of the graph.
You will want to continue to practice these skills over and over
until you're comfortable doing these examples and others like them comfortably.