Example 1 Sketch
the graph of each of the following parabolas.
(a) [Solution]
(b) [Solution]
(c) [Solution]
Solution
Okay, in all of these we will simply go through the
process given above to find the needed points and the graph.
(a)
First we need to find the vertex. We will need to be careful with the signs
however. Comparing our equation to the
form above we see that we must have and since that is the only way to get the
correct signs in our function.
Therefore, the vertex of this parabola is,
Now let’s find the yintercept. This is nothing more than a quick function
evaluation.
Next we need to find the xintercepts. This means
we’ll need to solve an equation.
However, before we do that we can actually tell whether or not we’ll
have any before we even start to solve the equation.
In this case we have which is positive and so we know that the
parabola opens up. Also the vertex is
a point below the xaxis. So, we know that the parabola will have at
least a few points below the xaxis
and it will open up. Therefore, since
once a parabola starts to open up it will continue to open up eventually we
will have to cross the xaxis. In other words, there are xintercepts for this parabola.
To find them we need to solve the following equation.
We solve equations like this back when we were solving quadratic equations so hopefully you
remember how to do them.
The two xintercepts
are then,
Now, at this point we’ve got points on either side of the
vertex so we are officially done with finding the points. However, let’s talk a little bit about how
to find a second point using the yintercept
and the axis of symmetry since we will need to do that eventually.
First, notice that the yintercept
has an x coordinate of 0 while the
vertex has an x coordinate of
3. This means that the yintercept is a distance of 3 to the
right of the axis of symmetry since that will move straight up from the
vertex.
Now, the left part of the graph will be a mirror image of
the right part of the graph. So, since
there is a point at that is a distance of 3 to the right of the
axis of symmetry there must also be a point at that is a distance of 3 to the left of the
axis of symmetry.
So, since the x
coordinate of the vertex is 3 and this new point is a distance of 3 to the
left its x coordinate must be
6. The coordinates of this new point
are then . We can verify this by evaluating the
function at . If we are correct we should get a value of
10. Let’s verify this.
So, we were correct.
Note that we usually don’t bother with the verification of this point.
Okay, it’s time to sketch the graph of the parabola. Here it is.
Note that we included the axis of symmetry in this graph
and typically we won’t. It was just
included here since we were discussing it earlier.
[Return to Problems]
(b)
Okay with this one we won’t put in quite a much
detail. First let’s notice that which is negative and so we know that this
parabola will open downward.
Next, by comparing our function to the general form we see
that the vertex of this parabola is . Again, be careful to get the signs correct
here!
Now let’s get the yintercept.
The yintercept
is then .
Now, we know that the vertex starts out below the xaxis and the parabola opens
down. This means that there can’t
possibly be xintercepts since the x axis is above the vertex and the
parabola will always open down. This
means that there is no reason, in general, to go through the solving process
to find what won’t exist.
However, let’s do it anyway. This will show us what to look for if we
don’t catch right away that they won’t exist from the vertex and direction
the parabola opens. We’ll need to
solve,
So, we got complex solutions. Complex solutions will always indicate no xintercepts.
Now, we do want points on either side of the vertex so
we’ll use the yintercept and the
axis of symmetry to get a second point.
The yintercept is a
distance of two to the left of the axis of symmetry and is at and so there must be a second point at the
same y value only a distance of 2
to the right of the axis of symmetry.
The coordinates of this point must then be .
Here is the sketch of this parabola.
[Return to Problems]
(c)
This one is actually a fairly simple one to graph. We’ll first notice that it will open
upwards.
Now, the vertex is probably the point where most students
run into trouble here. Since we have x^{2} by itself this means
that we must have and so the vertex is .
Note that this means there will not be any xintercepts with this parabola since
the vertex is above the xaxis and
the parabola opens upwards.
Next, the yintercept
is,
The yintercept
is exactly the same as the vertex.
This will happen on occasion so we shouldn’t get too worried about it
when that happens. Although this will
mean that we aren’t going to be able to use the yintercept to find a second point on the other side of the
vertex this time. In fact, we don’t
even have a point yet that isn’t the vertex!
So, we’ll need to find a point on either side of the
vertex. In this case since the
function isn’t too bad we’ll just plug in a couple of points.
Note that we could have gotten the second point here using
the axis of symmetry if we’d wanted to.
Here is a sketch of the graph.
[Return to Problems]
