In this section we want to look at the graph of a quadratic function.
The most general form of a quadratic function is,
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f (x) = ax2 + bx + c
All parabolas are vaguely “U” shaped and the will have
a highest or lowest point that is called the vertex. Parabolas may open
up or down and may or may not have x-intercepts and they will always
have a single y-intercept.
Note as well that a parabola that opens down will
always open down and a parabola that opens up will always open up. In other
words a parabola will not all of a sudden turn around and start opening up if it has already started opening down.
Similarly, if it has already started opening up it will not turn around and start
opening down all of a sudden.
The dashed line with each of these parabolas is called
the axis of symmetry. Every parabola has an axis of symmetry and, as the
graph shows, the graph to either side of the axis of symmetry is a mirror image
of the other side. This means that if we know a point on one side of the
parabola we will also know a point on the other side based on the axis of symmetry.
We will see how to find this point once we get into some examples.
We should probably do a quick review of intercepts before going much farther. Intercepts are the points where the
graph will cross the x or y-axis. We also saw a graph in the
section where we introduced intercepts where an intercept just touched the axis
without actually crossing it.
Finding intercepts is a fairly simple process. To find
the y-intercept of a function y = f (x) all we need to do is set x = 0 and evaluate to find the y coordinate. In
other words, the y-intercept is the point (0, f(0)) . We find x-intercepts in pretty much the same
way. We set y = 0 and
solve the resulting equation for the x coordinates.
So, we will need to solve the equation,
f (x) = 0
Now, let’s get back to parabolas. There is a basic
process we can always use to get a pretty good sketch of a parabola. Here it
is.
Sketching Parabolas
1. Find
the vertex. We’ll discuss how to find this shortly. It’s fairly simple, but
there are several methods for finding it and so will be discussed separately.
2. Find
the y-intercept, (0, f (0)) .
3. Solve
f (x) = 0 to find the x
coordinates of the x-intercepts if they exist. As we will see in our
examples we can have 0, 1, or 2 x-intercepts.
4. Make
sure that you’ve got at least one point to either side of the vertex. This is
to make sure we get a somewhat accurate sketch. If the parabola has two x-intercepts
then we’ll already have these points. If it has 0 or 1 x-intercept we
can either just plug in another x value or use the y-intercept
and the axis of symmetry to get the second point.
5. Sketch
the graph. At this point we’ve gotten enough points to get a fairly decent idea
of what the parabola will look like.
Now, there are two forms of the parabola that we will be looking
at. This first form will make graphing parabolas very easy. Unfortunately, most parabolas are not
in this form. The second form is the more common form and will require slightly (and only
slightly) more work to sketch the graph of the parabola.
Let’s take a look at the first form of the parabola.
f (x) = a (x – h)2 + k
There are two pieces of information about the parabola that we can
instantly get from this function. First, if a is positive then the parabola will
open up and if a is negative then the parabola will open down. Secondly, the vertex of the parabola is
the point (h, k ) . Be
very careful with signs when getting the vertex here.
So, when we are lucky enough to have this form of the parabola we
are given the vertex for free.
Okay, we’ve seen some examples now of this form of the parabola.
However, as noted earlier most parabolas are not given in that form. So, we
need to take a look at how to graph a parabola that is in the general form.
f (x) = ax2 + bx + c
In this form the sign of a will determine whether or not the
parabola will open upwards or downwards just as it did in the previous set of
examples. Unlike the previous form we will not get the vertex for free this
time. However, it is will easy to find. Here is the vertex for a parabola in the general form.
To get the vertex all we do is compute the x coordinate from
a and b and then plug this into the function to get the y coordinate.
Not quite as simple as the previous form, but still not all that difficult.
Note as well that we will get the y-intercept for free from
this form. The y-intercept is,
f (0) = a (0)2 + b (0) + c = c ⇒ (0, c)
so we won’t need to do any computations for this one.
As a final topic in this section we need to briefly
talk about how to take a parabola in the general form and convert it into the
form
f (x) = a (x − h)2 + k
This will use a modified completing the square process.
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