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, .
   
   
3. Solve  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.

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 y-intercept.  This is nothing more than a quick function evaluation.

Next we need to find the x-intercepts.  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 x-axis.  So, we know that the parabola will have at least a few points below the x-axis 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 x-axis.  In other words, there are x-intercepts 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 x-intercepts 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 y-intercept and the axis of symmetry since we will need to do that eventually.

First, notice that the y-intercept has an x coordinate of 0 while the vertex has an x coordinate of -3.  This means that the y-intercept 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.

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(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 y-intercept.

The y-intercept is then .

Now, we know that the vertex starts out below the x-axis and the parabola opens down.  This means that there can’t possibly be x-intercepts 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 x-intercepts.

Now, we do want points on either side of the vertex so we’ll use the y-intercept and the axis of symmetry to get a second point.  The y-intercept 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.

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(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 x² by itself this means that we must have  and so the vertex is .

Note that this means there will not be any x-intercepts with this parabola since the vertex is above the x-axis and the parabola opens upwards.

Next, the y-intercept is,

The y-intercept 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 y-intercept 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.

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Example 2  Sketch the graph of each of the following parabolas.

(a)    [Solution]

(b)    [Solution]

(c)    [Solution]

Solution

(a) For this parabola we’ve got ,  and .  Make sure that you’re careful with signs when identifying these values.  So we know that this parabola will open up since a is positive. 

Here are the evaluations for the vertex.

The vertex is then .  Now at this point we also know that there won’t be any x-intercepts for this parabola since the vertex is above the x-axis and it opens upward.

The y-intercept is  and using the axis of symmetry we know that  must also be on the parabola.

Here is a sketch of the parabola.

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(b) In this case ,  and .  From these we see that the parabola will open downward since a is negative.  Here are the vertex evaluations.

So, the vertex is  and we also can see that this time there will be x-intercepts.  In fact, let’s go ahead and find them now.

So, the x-intercepts are  and .  Notice that  is also the y-intercept.  This will happen on occasion so don’t get excited about it when it does.

At this point we’ve got all the information that we need in order to sketch the graph so here it is,

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(c)  In this final part we have ,  and .  So, this parabola will open up.

Here are the vertex evaluations.

So, the vertex is .  Note that since the y coordinate of this point is zero it is also an x-intercept.  In fact it will be the only x-intercept for this graph.  This makes sense if we consider the fact that the vertex, in this case, is the lowest point on the graph and so the graph simply can’t touch the x-axis anywhere else.

The fact that this parabola has only one x-intercept can be verified by solving as we’ve done in the other examples to this point.

Sure enough there is only one x-intercept.  Note that this will mean that we’re going to have to use the axis of symmetry to get a second point from the y-intercept in this case.

Speaking of which, the y-intercept in this case is .  This means that the second point is .

Here is a sketch of the graph.

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Example 3  Convert each of the following into the form .

(a)    [Solution]

(b)    [Solution]

Solution

Okay, as we pointed out above we are going to complete the square here.  However, it is a slightly different process than the other times that we’ve seen it to this point.

(a) The thing that we’ve got to remember here is that we must have a coefficient of 1 for the x² term in order to complete the square.  So, to get that we will first factor the coefficient of the x² term out of the whole right side as follows.

Note that this will often put fractions into the problem that is just something that we’ll need to be able to deal with.  Also note that if we’re lucky enough to have a coefficient of 1 on the x² term we won’t have to do this step.

Now, this is where the process really starts differing from what we’ve seen to this point.  We still take one-half the coefficient of x and square it.  However, instead of adding this to both sides we do the following with it.

We add and subtract this quantity inside the parenthesis as shown.  Note that all we are really doing here is adding in zero since 9-9=0!  The order listed here is important.  We MUST add first and then subtract.

The next step is to factor the first three terms and combine the last two as follows.

As a final step we multiply the 2 back through.

And there we go.

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(b) Be careful here.  We don’t have a coefficient of 1 on the x² term, we’ve got a coefficient of -1.  So, the process is identical outside of that so we won’t put in as much detail this time.

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