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### Three Ways to Find the Vertex of a Parabola

```Date: 02/19/2007 at 15:43:14
From: Jackie
Subject: H(x) = -x2-8x-15

For H(x) = -x2-8x-15, what are the coordinates of the vertex, what is
the equation of the axis of symmetry, what is the greatest value of
the function?

```

```
Date: 02/19/2007 at 18:55:45
From: Doctor Greenie
Subject: Re: H(x) = -x2-8x-15

Hi, Jackie--

There are several ways to solve this type of problem.  Let me show
you the detailed solution to your problem by three different methods.
If you want to do well on this topic--and in future math classes,
where this type of problem comes up frequently--you will do well to
try to understand all of them.

Before we start, note that the coordinates of the vertex of the
parabola exactly determine the equation of the line of symmetry and
the maximum or minimum value.  So answering all the questions is a
matter of finding the coordinates of the vertex of the parabola.

First method: completing the square and vertex form of the equation

The "vertex" form of the equation of a parabola is

y = a(x - h)^2 + k

where the vertex is at the point (h,k).  So we want to take the given
equation

y = -x^2 - 8x - 15

and put it in vertex form.  We do this by completing the square, so
that the part of the equation involving x can be written in the form
(x - h)^2.  In completing the square, we are only going to use the x^2
and x terms of the equation, so we are going to leave the constant
term alone for now.  And we are going to factor out the coefficient
of the x^2 term, to get the "x^2" by itself.  When we do these two
things, our equation looks like this:

y = -1(x^2 + 8x   ) - 15

Again, note that we have left the constant term "-15" alone, and we
have factored the coefficient "-1" out of the x^2 and x terms of the
equation.

Now, inside the parentheses, we need to add a number which makes the
whole expression a perfect square--of the desired form (x - h)^2.  The
rule here is to take half the coefficient of the "x" term and square
it.  Half of 8 is 4; 4 squared is 16, so we add 16 to the expression
in parentheses:

y = -1(x^2 + 8x + 16) - 15 (this equation is incomplete; see below)

Now, we can't just randomly add something to our equation; we need it
to stay unchanged when we complete the square.  We have to add
something outside the parentheses that offsets the change we made
inside the parentheses.  We added "16" inside the parentheses; but
the entire expression inside the parentheses is multiplied by -1, so
we really added -16 to the equation.  To offset that, we need to add
+16 outside the parentheses.  The equation we get is this:

y = -1(x^2 + 8x + 16) - 15 + 16

Again, note that we have added -1(16) = -16 inside the parentheses;
to keep the entire expression unchanged, we have also added +16
outside the parentheses.

Now we write the perfect square in the parentheses in its desired form
and simplify the rest of the equation:

y = -1(x + 4)^2 + 1

Remember that the vertex form of the equation is

y = a(x - h)^2 + k

So we know that a = -1, which means the parabola opens downward; that
in turn means the vertex is the maximum point of the parabola.  And
note that our equation shows "(x + 4)^2", while the vertex form
shows "(x - h)^2".  That means that

x - h = x + 4

and so

h = -4

And so, according to vertex form, our vertex (h,k) is (-4,1).  And
then we know the line of symmetry is x = -4, and the maximum value of
the equation is 1, which is the y-value of the vertex.

We have done the problem using the method of completing the square to
put the equation in vertex form.

The next two processes I am going to show are shortcuts to the
answer to the problem.  Many students memorize these shortcuts and
make no effort to understand the above process; this puts them at a
great disadvantage if they go on to further math classes.  I
strongly recommend that you both learn the shortcuts (for speed) but
also learn the above process; doing so will give you a better
understanding of the whole topic, and it will put you in a better
position for continuing your math studies.

Second method: factoring and using symmetry

In your example, the equation factors nicely.  We can use this and
the known fact that the parabola has a line of symmetry to get to
the solution much faster than by completing the square.

y = -x^2 - 8x - 15
y = -1(x^2 + 8x + 15)  [factor out the coefficient of the x^2 term]
y = -1(x + 3)(x + 5)

This factored form tells us that the equation has the value of zero
when x + 3 = 0 (i.e., x = -3) and when x + 5 = 0 (i.e., x = -5).  So
the parabola crosses the x-axis at -3 and -5.  But since the parabola
is symmetric, the line of symmetry must be halfway between these two
points--at x = -4.  So the x-coordinate of the vertex is -4.

To find the y-value of the vertex, we simply substitute x = -4 in the
equation.  It is probably easiest to do this in the final factored
form shown above:

y = -1(-4 + 3)(-4 + 5) = -1(-1)(1) = 1

So again the coordinates of the vertex are (x,y) = (-4,1).

Again, let me emphasize that this method is much faster than
completing the square but it doesn't always work (the parabola might
not cross the x-axis).  And you will have a much better understanding
of this topic if you know the completing the square method.

Third method: a quick and easy formula

If the equation of the parabola is written in the form

y = ax^2 + bx + c

then the x-coordinate of the vertex is given by the easy formula

x = -b/2a

In your example, the equation is

y = -x^2 - 8x - 15

So a = -1 and b = -8; then the formula for the x-coordinate of the
vertex gives us

x = -(-8)/2(-1) = 8/(-2) = -4

And again we substitute this value of x into the equation to find the
corresponding value of y:

y = -(-4)^2 -8(-4) - 15 = -16 + 32 - 15 = 1

This method is by far the fastest and easiest.  And you SHOULD use
it whenever you are allowed to, to make less work for yourself.  But
don't fall into the trap of just memorizing this quick method and
ignoring the much more important method of completing the square.

I hope all this helps rather than confuses.  Please write back if
you have any further questions about any of this.

Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles

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