Solution to Problem 9.7

7) The problem is: Find the fourth degree (quartic) polynomial that passes through the origin and to which the line

   x + 2y  =  14

is tangent at both x = 4 and at x = -2.

Your first hint was an easy one -- put the line into slope-intercept form. That would be

           x
   y  =  -   + 7
           2

So the slope of the line is m = -1/2. What is so significant about this slope? It is that the derivative of the polynomial you are looking for must be equal to this slope at each of the two points of tangency. The hint also suggested that you find the y value of the line at each of the two x values given in the problem. Again that's pretty easy:

              4
   y(4)  =  -   + 7  =  5
              2

             2
   y(-2)  =    + 7  =  8
             2

Now we get to the hard part, which is establishing the five coefficients of the fourth degree polynomial that meets the criteria given in the problem. One of the criteria is that the polynomial pass through the origin. Only polynomials that have zero for a constant term can ever pass through the origin. So that tells you one of the five coefficients. That is, the constant coefficient (the one that has no x multiplied by it) must be zero. Writing the polynomial using the remaining four unknown coefficients gives:

   P(x)  =  Ax⁴ + Bx³ + Cx² + Dx

Taking the derivative of this gives

   P'(x)  =  4Ax³ + 3Bx² + 2Cx + D

For the line to be tangent to the polynomial at x = 4, the line must be equal to the polynomial at that point. So P(4) = y(4). Putting the numbers in for x, that equation becomes:

   P(4)   =   256A  +  64B  +  16C  +  4D   =   5

Likewise you must also have P(-2) = y(-2). That translates to

   P(-2)  =  16A  -  8B  +  4C  -  2D  =  8

For the line to be tangent to P(x) at both the points specified in the problem, P'(x) = m at each of those points. That translates to

                                             1
   P'(4)  =  256A  +  48B  +  8C  +  D  =  -    =  -0.5
                                             2

and

                                              1
   P'(-2)  =  -32A  +  12B  -  4C  +  D  =  -    =  -0.5
                                              2

That gives you four linear equations in the four unknowns, A, B, C, and D.

Solve them either by hand, or using the linear equation solver. You will get (to 4 figures beyond the decimal):

   A  =  -0.1094

   B  =   0.4375

   C  =   1.3125

   D  =  -4.0000

So the polynomial is

   P(x)  =  -0.1094x⁴ + 0.4375x³ + 1.3125x² - 4.0000x

Here is a plot of that polynomial and the line, x + 2y = 14, showing the tangency at the two points.

Just for grins, see what happens when you try to find the third degree (cubic) polynomial to which the line given in problem 7 is tangent at x = 4 and x = -2 (dropping the requirement that the polynomial pass through the origin). You should find that the cubed term and the squared term are both zero. What do you see about the remaining terms? What does that tell you about the existence of a genuine cubic (or quadratic for that matter) that meets the criteria?

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