EMAT 4680

31 October 2000

Homework 16

1.) The following functions are plotted on the same axes (function colors are in parentheses):

        y = x + 4 (red)         y = x + 3 (blue)         y = x + 2 (magenta)        y = x + 1 (maroon)

        y = x (green)             y = x - 1 (violet)        y = x - 2 (cyan)             y = x - 3 (coral)         y = x - 4 (navy)

The basic form of a linear equation is y = x. Let f(x) = y + b. The above graph illustrates that including the value b in the function shifts the y-intercept of f(x) b places from the y-intercept of y = x, which is 0. Therefore, the x-intercept shifts -b places from the x-intercept of y = x, which is 0. This graphing activity would be meaningful in a high school classroom because it would show students the transformation of y = x by b into the function f(x) = y + b. The students could generalize this transformation for any initial, linear y equation.

2.) The following functions are plotted on the same of axes (functions colors are in parentheses):

        y = (x - 2) + 4 (red)         y = (x - 2) + 3 (blue)         y = (x - 2) + 2 (magenta)        y = (x - 2) + 1 (maroon)

        y = (x - 2) + 0 (green)     y = (x - 2) - 1 (violet)        y = (x - 2) - 2 (cyan)                 y = (x - 2) - 3 (coral)
 
        y = (x - 2) - 4 (navy)

As discussed in problem 1, f(x) = y + b. However, the original function y now states y = x - 2 instead of y = x. The original graph has a y-intercept at -2 and an x-intercept at 2. For f(x) = y + b, the original y-intercept is shifted b spaces on the y-axis. Additionally, the original x-intercept is shifted -b spaces on the x-axis. This exercise would be meaningful in a high school classroom just as the first exercise would be meaningful to illustrate the translation of the original linear function by b.

3.) The following functions are plotted on the same axes. (Function colors are in parentheses):

        y = 4(x - 2) (red)         y = 3(x - 2) (blue)         y = 2(x - 2) (magenta)        y = x - 2 (maroon)
 
        y = 0(x - 2) (green)     y = -1(x - 2) (violet)        y = -2(x - 2) (cyan)          y = -3(x - 2) (coral)

        y = -4(x - 2) (navy)

For this exercise the original function is y = x - 2. All other functions are of the form f(x) = by. The original function has a slope of 1, a y-intercept of -2, and an x-intercept of 2. For f(x), the slope is b, the y-intercept is -2b, and the x-intercept is always 2. This exercise would be meaningful in a high school classroom because it shows that translation of y = any linear function to f(x) = by.

4.) The following functions are plotted on the same axes. (Function colors are in parentheses):

        y = x^2 + 3 (red)         y = x^2 + 2 (blue)         y = x^2 + 1 (magenta)        y = x^2 + 0 (maroon)

        y = x^2 - 1 (green)      y = x^2 - 2 (violet)        y = x^2 - 3 (cyan)

The original function in this exercise is y = x^2 + 0 = x^2. All other functions are of the form f(x) = y + b. The vertex for the original function is (0,0). Therefore, the y-intercept is 0, and the x-intercept is 0. The vertex for f(x) is (0,b). Therefore, the y-intercept is b. The number of x-intercepts increases when b is negative. Thus the number of real solutions to f(x) increases from 1 to 2 when b is negative. In return, the number of x-intercepts decreases when b is positive. Thus the number of real solutions to f(x) decreases from 1 to 0 when b is positive. This exercise would be meaningful in the high school classroom to illustrate the effect of transforming y = any quadratic function to f(x) = y + b.

5.) The following functions are plotted on the same axes. (Function colors are in parentheses):

        y = (x + 3)^2 (red)         y = (x + 2)^2 (blue)         y = (x + 1)^2 (magenta)         y = (x + 0)^2 (maroon)

        y = (x - 1)^2 (green)      y = (x - 2)^2 (violet)        y = (x - 3)^2 (cyan)