Graphing and Common Graphs
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Sketch the graph of each of the following.
This is a line in the slope
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
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.
This is a parabola in the form
Parabolas in this form will have
the vertex at (h, k). So, the vertex for our parabola is
(-3,-1). We can also notice that this
parabola will open up since the coefficient of the squared term is positive
(and of course it would open down if it was negative).
In graphing parabolas it is also
convenient to have the x-intercepts,
if they exist. In this case they will
exist since the vertex is below the x-axis
and the parabola opens up. To find them
all we need to do is solve.
With this information we can plot
the main points and sketch in the parabola.
Here is the sketch.
This is also the graph of a
parabola only it is in the more general form.
In this form, the x-coordinate of the vertex is and we get the y-coordinate by plugging this value back into the equation. So, for our parabola the coordinates of the
vertex will be.
So, the vertex for this parabola
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.
This also means that we’ll have x-intercepts on this graph. So, we’ll solve the following.
So, we will have x-intercepts at and . Notice that to make my life easier in the
solution process I multiplied everything by -1 to get the coefficient of the positive.
This made the factoring easier.
Here’s a sketch of this parabola.
We could also use completing the
square to convert this into the form used in the previous problem. To do this I’ll first factor out a -1 since
it’s always easier to complete the square if the coefficient of the is a positive 1.
Now take half the coefficient of
the x and square that. Then add and subtract this to the quantity
inside the parenthesis. This will make
the first three terms a perfect square which can be factored.
The final step is to multiply the minus sign back
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 .
is a parabola that opens up and
has a vertex of (3,-4).
Well our function is in the form
and this is a parabola that opens
left or right depending on the sign of a
(right if a is positive and left if a is negative). The y-coordinate
of the vertex is given by and we find the x-coordinate by plugging this into the equation.
Our function is a parabola that
opens to the right (a is positive)
and has a vertex at
(-4,3). To graph this we’ll need y-intercepts. We find these
just like we found x-intercepts in
the previous couple of problems.
The parabola will have y-intercepts at and . Here’s a sketch of the graph.
This is a circle in it s standard
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 (0, -5)
and a radius of 2. Here’s a sketch of
This is also a circle, but it
isn’t in standard form. To graph this
we’ll need to do it put it into standard form.
That’s easy enough to do however.
All we need to do is complete the square on the x’s and on the y’s.
So, it looks like the center is
(-1, 4) and the radius is 3. Here’s a
sketch of the circle.
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 . Here’s a sketch of the ellipse.
This is a hyperbola. There are actually two standard forms for a
Opens right and left
Opens up and down
a units right
b units up and
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 hyperbola opens up and down.
Both have the same “center”. 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 up 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 they asymptotes.
For the equation listed here the
hyperbola will open left and right. Its
(-1, 2). The two vertices are (-4, 2) and (2, 2). The asymptotes will have slopes .
Here is a sketch of this
There isn’t much to this problem. This
is just the square root and it’s a graph that you need to be able to sketch on
occasion so here it is. Just remember
that you can’t plug any negative x
into the square root. Here is the graph.
Another simple graph that doesn’t really need any discussion, it is here simply
to make sure you can sketch it.
This last sketch is not that difficult if you remember how to evaluate Absolute Value functions. Here is the sketch for this one.