The General Polynomial Function

Date: 09/17/98 at 22:55:29
From: prayoon Chaithaweep
Subject: Pre-calculus

Can you tell me how to solve these problems:

1) For the general polynomial function:

          n      n-1      n-2 
   y = A x  + A x    + A x    + ... + A   
        0        1        2            n

what are the corresponding values of A , A , A , . . . A  , and n in                                  
                                      0   1   2         n
the linear function y = 6x - 2 ?

2) Write a linear function for A = 2/3, A = 7, and n = 1:
                                0        1

I know the answer of 2 is y = 2/3 + 7, but I don't know how to find it.

Date: 09/18/98 at 12:25:10
From: Doctor Peterson
Subject: Re: Pre-calculus

Hello, Prayoon. I've had to reformat your question because the spacing 
got messed up,  but I think I got it right. It's easier to write the 
subscripts as "A_0" for A-subscript-0, and "x^n" for x-to-the-nth 
power.

They've made it a little tricky by using A_0 for the coefficient of 
the x^n term, but it's not too hard. First, n will be the highest 
exponent, so in problem 1, n = 1. Then you can just write the general 
form for n = 1, and write your given equation under it, remembering 
that x^0 = 1 and x^1 = x:

   y = A_0 x^1 + A_1 x^0
   y =  6  x^1 -  2  x^0

So A_0 is 6 and A_1 is -2, since 6x - 2 can be written as:

   6x^1 + (-2)x^0

The second problem just reverses the process. You are given n = 1, so 
you can write the generic form again, then fill in the values:

   y = A_0 x^1 + A_1 x^0
   y = 2/3 x^1 +  7  x^0
     = 2/3 x   +  7

I suspect that your problem may be that you are not quite comfortable 
with what x^1 and x^0 are. I hope this helps you.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   

Date: 09/18/98 at 17:20:51
From: Anonymous
Subject: Re: Pre-calculus

I still don't understand why n = 1 in problem 1.
Also, how do I graph this function f(x) = square root 2x and how I 
find the inverse of it?

Prayoon

Date: 09/19/98 at 18:00:57
From: Doctor Peterson
Subject: Re: Pre-calculus

Hello, Prayoon. 

For further explanation of the first problem, in your formula for a 
polynomial, n is the highest exponent of x in the polynomial. In 
6x - 2, which can be rewritten as 6x^1 - 2x^0, you have two powers of 
x, x = x^1 and 1 = x^0. So the highest exponent is 1.

You want to graph y = sqrt(2x). You could just choose some values for 
x and plot them the way you make any graph, but let's approach this in 
a way that is connected with inverses.

Your equation is the same as y^2 = 2x, or y^2/2 = x, with the 
restriction that y has to be positive, since the square root gives 
only the positive number whose square is 2x. So we can graph it by 
drawing the parabola x = y^2/2, looking at our paper sideways. Here's 
y = x^2 / 2, the same parabola going the other way:

              |
      *       +       *
              |
              +
          *   |   *
   ---+---+---*---+---+---
     -2  -1   0   1   2

and here is x = y^2 / 2 (in which I have used only the positive values 
for y):

              |
              +       *
              |
              + *
              |    
   ---+---+---*---+---+---
     -2  -1   0   1   2

To find the inverse, you have to solve for x in terms of y and state 
the restriction on the function's domain. That's really what we just 
did: x = y^2/2, with y positive. So the inverse of f is:

   f(y) = y^2/2   for y >= 0

(You could also write this as:

   f(x) = x^2/2   for x >= 0

since a function doesn't care what you call its variable.)

I hope that gives you an idea of how inverses are related to graphs.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/