

The purpose of this section is to make sure that you’re
familiar with the graphs of many of the basic functions that you’re liable to
run across in a calculus class.
Example 1 Graph
.
Solution
This is a line in the slope intercept form
In this case the line has a y intercept of (0,b)
and a slope of m. Recall that slope can be thought of as
Note that if the slope is negative we tend to think of the
rise as a fall.
The slope allows us to get a second point on the
line. Once we have any point on the
line and the slope we move right by run
and up/down by rise depending on
the sign. This will be a second point
on the line.
In this case we know (0,3) is a point on the line and the
slope is . So starting at (0,3) we’ll move 5 to the
right (i.e. ) and down 2 (i.e. ) to get (5,1) as a second point on
the line. Once we’ve got two points on
a line all we need to do is plot the two points and connect them with a line.
Here’s the sketch for this line.

Example 2 Graph
Solution
There really isn’t much to this problem outside of
reminding ourselves of what absolute value is. Recall that the absolute value function is
defined as,
The graph is then,

Example 3 Graph .
Solution
This is a parabola in the general form.
In this form, the xcoordinate
of the vertex (the highest or lowest point on the parabola) is and we get the ycoordinate is . So, for our parabola the coordinates of the
vertex will be.
So, the vertex for this parabola is (1,4).
We can also determine which direction the parabola opens
from the sign of a. If a
is positive the parabola opens up and if a
is negative the parabola opens down.
In our case the parabola opens down.
Now, because the vertex is above the xaxis and the parabola opens down we know that we’ll have xintercepts (i.e. values of x for
which we’ll have ) on this graph. So, we’ll solve the following.
So, we will have xintercepts
at and . Notice that to make our life easier in the
solution process we multiplied everything by 1 to get the coefficient of the
positive.
This made the factoring easier.
Here’s a sketch of this parabola.

Example 4 Graph
Solution
Most people come out of an Algebra class capable of
dealing with functions in the form . However, many functions that you will have
to deal with in a Calculus class are in the form and can only be easily worked with in that
form. So, you need to get used to
working with functions in this form.
The nice thing about these kinds of function is that if
you can deal with functions in the form then you can deal with functions in the form
even if you aren’t that familiar with them.
Let’s first consider the equation.
This is a parabola that opens up and has a vertex of
(3,4), as we know from our work in the previous example.
For our function we have essentially the same equation
except the x and y’s are switched around. In other words, we have a parabola in the
form,
This is the general form of this kind of parabola and this
will be a parabola that opens left or right depending on the sign of a.
The ycoordinate of the
vertex is given by and we find the xcoordinate by plugging this into the equation. So, you can see that this is very similar
to the type of parabola that you’re already used to dealing with.
Now, let’s get back to the example. Our function is a parabola that opens to
the right (a is positive) and has a
vertex at (4,3). The vertex is to the
left of the yaxis and opens to the
right so we’ll need the yintercepts
(i.e. values of y for which we’ll have )).
We find these just like we found xintercepts
in the previous problem.
So, our parabola will have yintercepts at and . Here’s a sketch of the graph.

Example 5 Graph
.
Solution
To determine just what kind of graph we’ve got here we
need to complete the square on both the x
and the y.
Recall that to complete the square we take the half of the
coefficient of the x (or the y), square this and then add and subtract
it to the equation.
Upon doing this we see that we have a circle and it’s now
written in standard form.
When circles are in this form we can easily identify the
center : (h, k) and radius : r.
Once we have these we can graph the circle simply by starting at the
center and moving right, left, up and down by r to get the rightmost, leftmost, top most and bottom most points
respectively.
Our circle has a center at (1, 4) and a radius of 3. Here’s a sketch of this circle.

Example 6 Graph
Solution
This is an ellipse.
The standard form of the ellipse is
This is an ellipse with center (h, k) and the right most and left most points are a distance of a away from the center and the top
most and bottom most points are a distance of b away from the center.
The ellipse for this problem has center (2, 2) and has and . Note that to get the b we’re really rewriting the equation as,
to get it into standard from.
Here’s a sketch of the ellipse.

Example 7 Graph
Solution
This is a hyperbola.
There are actually two standard forms for a hyperbola. Here are the basics for each form.
Form






Center

(h, k)

(h, k)




Opens

Opens right and left

Opens up and down




Vertices

a units right
and left
of center.

b units up and
down
from center.




Slope of Asymptotes



So, what does all this mean? First, notice that one of the terms is
positive and the other is negative.
This will determine which direction the two parts of the hyperbola
open. If the x term is positive the hyperbola opens left and right. Likewise, if the y term is positive the parabola opens up and down.
Both have the same “center”. Note that hyperbolas don’t really have a
center in the sense that circles and ellipses have centers. The center is the starting point in
graphing a hyperbola. It tells us how
to get to the vertices and how to get the asymptotes set up.
The asymptotes of a hyperbola are two lines that intersect
at the center and have the slopes listed above. As you move farther out from the center the
graph will get closer and closer to the asymptotes.
For the equation listed here the hyperbola will open left
and right. Its center is
(1, 2). The two
vertices are (4, 2) and (2, 2). The
asymptotes will have slopes .
Here is a sketch of this hyperbola. Note that the asymptotes are denoted by the
two dashed lines.

Example 8 Graph
and .
Solution
There really isn’t a lot to this problem other than making
sure that both of these exponentials are graphed somewhere.
These will both show up with some regularity in later
sections and their behavior as x
goes to both plus and minus infinity will be needed and from this graph we
can clearly see this behavior.

Example 9 Graph
.
Solution
This has already been graphed once in this review, but
this puts it here with all the other “important” graphs.

Example 10 Graph
.
Solution
This one is fairly simple, we just need to make sure that
we can graph it when need be.
Remember that the domain of the square root function is .

Example 11 Graph
Solution
Again, there really isn’t much to this other than to make
sure it’s been graphed somewhere so we can say we’ve done it.

Example 12 Graph
Solution
There really isn’t a whole lot to this one. Here’s the graph for .
Let’s also note here that we can put all values of x into cosine (which won’t be the case
for most of the trig functions) and so the domain is all real numbers. Also
note that
It is important to notice that cosine will never be larger
than 1 or smaller than 1. This will
be useful on occasion in a calculus class.
In general we can say that

Example 13 Graph
Solution
As with the first problem in this section there really
isn’t a lot to do other than graph it.
Here is the graph.
From this graph we can see that sine has the same range
that cosine does. In general
As with cosine, sine itself will never be larger than 1
and never smaller than 1. Also the
domain of sine is all real numbers.

Example 14 Graph
.
Solution
In the case of tangent we have to be careful when plugging
x’s in since tangent doesn’t exist
wherever cosine is zero (remember that ).
Tangent will not exist at
and the graph will have asymptotes at these points. Here is the graph of tangent on the range .


