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Quadratic Formulas, Equations, Parabolas, Graphing


Date: 6/16/96 at 22:30:33
From: Glen McKenzie
Subject: Quadratic Problem

My name is Brooke McKenzie.  I'm in grade ten math and I was wondering 
if you could help me with a couple problems.

We are doing Quadratic Formulas, Equations, and Parabolas, and Graphing 
them.

The first equation is:

6x(squared) + x - 1 = 0  (I'm not sure how to say that just the x is 
squared)

*I need to find the roots (x-intercepts), the vertex, some coordinates 
and graph it.*

I am mostly having problems graphing because I end up with numbers 
like 1 1/24, -1/12, 14 and so on.  I'll show you how I worked it out and 
maybe you can see if I'm doing something wrong or whatever.

FINDING ROOTS

6x^2 + x = 0
(2x + 1)(3x - 1)
2x+1=0		3x-1=0
x= -1/2		x= 1/3

roots (-1/2, 0)  (1/3, 0)


FINDING VERTEX

m=  -1/2 + 1/3      y=6 (x squared) + x - 1
     ---------       =6(-1/12 squared) + (-1/12) -1
	      2      =6 (1/144) - 1/12 - 1	
=    3/6 + 2/6       =6/144 - 1/12 - 1
     ---------       =1/24 - 2/24 - 24/24
          2          =-1/24 - 24/24
=    -1/6            =-25/24
    ------           = -1 1/24
       2
=-1/6 x 1/2         vertex (-1/12, -1 1/24)

=-1/12

FINDING COORDINATES

these are what I used:

x          y

1          6
3/2       14
0         -1
-1         4
-3/2      11

I'll just do one to show you how I figured out 
the y coordinates.

6 (x squared) + x - 1 = y
6 (1 squared) + 1 - 1 = y
6 (1) + 1 - 1 = y
6 + 1 -	1 = y
6 = y


If you can see anything wrong or have any solutions to 
help me graph this I would really appreciate it.

Brooke


Date: 6/17/96 at 9:0:10
From: Doctor Anthony
Subject: Re: Quadratic Problem

>roots (-1/2, 0)  (1/3, 0)
The roots are (-1/2) and (1/3).  Do NOT include the 0's


>FINDING VERTEX
You can find the vertex without first finding the roots.  If the roots 
are irrational or non-existent, then you need an alternative approach 
anyhow.

6x^2 + x - 1 = y  First make the coefficient of x^2 equal to 1 by 
dividing through by 6.

  x^2 + (1/6)x - 1/6 = y/6

Now complete the square on x by adding (and later subtracting) the 
square of half the coefficient of x.  The coefficient of x is 1/6, so 
halve it = 1/12 and then square it = 1/144

  x^2 + (1/6)x + 1/144 -1/6 - 1/144 = y/6

    (x + 1/12)^2 - 25/144 = y/6

     
This reaches its minimum value when the bracket is zero (since 
anything squared will be positive).  The bracket is zero if x = -1/12, 
and in this case y/6 = -25/144, so y = -25/24

Vertex is at (-1/12,  -25/24)


>FINDING COORDINATES

If I were you, I would make up a table as shown below to plot the 
graph of the function:

   x| -2  | -1 | 0  |  1  |  2  |  3  | 
----------------------------------------
6x^2| 24  |  6 | 0  |  6  |  24 |  54 |  
  x | -2  | -1 | 0  |  1  |   2 |   3 |
 -1 | -1  | -1 | -1 | -1  |  -1 |  -1 |
----------------------------------------
 y= | 21  |  4 | -1 |  6  |  25 |  56 |

You can of course plot at closer intervals if you want to, and for a 
different range of values of x, but a table layout as I have shown it 
above will make the calculations easier to follow.

-Doctor Anthony,  The Math Forum
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Associated Topics:
High School Basic Algebra
High School Equations, Graphs, Translations
High School Polynomials

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