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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 

  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 

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 
Associated Topics:
High School Conic Sections/Circles

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