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Graphing Parabolas


Date: 01/16/97 at 23:59:49
From: Anonymous
Subject: Graphing Parabolas

I have trouble plotting points for a parabola.  I find points from 
putting the original equation into standard form.  I have trouble 
finding the vertex and the line of symmetry.  I need a trick or a
short-cut or something.

Thank you.


Date: 01/18/97 at 23:09:58
From: Doctor Scott
Subject: Re: Graphing Parabolas

Hi!

To find the vertex of a parabola, you probably learned a process 
called "completing the square" (shown below) which puts a quadratic 
function (a parabola) into standard form : y = a(x - b)^2 + c, where 
a, b, and c are constants. Once you have the equation in standard 
form, the vertex is at the point (b, c) and the line of symmetry is 
x = b. If a is positive, the parabola opens up, if a is negative, the 
parabola opens down.

If the original equation is x = ay^2 + by + c, then it is a parabola 
which lies "on its side" and opens either to the left or the right.  
You can use completing the square to put the equation in standard 
form, x = c(y - h)^2 + k, with its vertex at (k, h) and line of 
symmetry y = h.  

To plot the parabola, I usually find that it is easiest to create a 
t-table (or a table of values) using a few points on either side of 
the vertex to find some points which lie on the parabola.

Example:   y = 4x^2 + 8x + 6

Complete the square:    y - 6 = 4(x^2 + 2x   )    <- move the constant 
                                                     and factor out a 4

Now find half the "x" term and square it, adding this to the right 
(inside the parentheses) and keeping the equation in balance by 
putting the appropriate amount on the left side. So, add 1 to the 
right, but put 4 on the left (because of the 4 outside the 
parentheses):  
                                                                         
        y - 6 + 4 = 4(x^2 + 2x + 1)

The right side is now a perfect square trinomial - so it factors to 
(x + 1)^2:

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

Finally, isolate y:           

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

So, the vertex of the parabola is (-1, 2), the line of symmetry is 
x = -1 and to plot the parabola, create a small t-table:

         x   |   y
        -----------
        -3   |
        -2   |
        -1   |   2
         0   |
         1   |

Now, find the y-values for the table using the final form of the 
equation.  (Note that we chose 2 points on the left of the vertex and 
2 on the right).

Hope this helps you out!
           
-Doctor Scott,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Equations, Graphs, Translations

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