1201

Run Multiple Choice Generator at h:\shares\physics
Run Core Problems by giving the program 174 as your 'course'--enter this when asked for your 3-digit course number.
Do the 'quizzes' on Exponential Functions: Numbers and Exponential Functions: Graphs.
If you miss a problem be sure to click on Comment and explain what you do and do not understand about the question and the given solution

http://www.vhcc.edu/dsmith/images/graph_series_1.htm

On the given graph we have an increasing linear function with an x-intercept to the left of the origin, and a decreasing linear function with an x-intercept to the right of the origin.

We wish to construct a graph of g / f.

When g = 0 we have g / f = 0 (caution: this is so provided g is not also 0; if f and g are both zero at a point we get 0 / 0, which is not defined).
When f = 0 we have g / f undefined, because no matter what g is, dividing it by 0 is not defined.
When f = 1 we have g / f = g, so at the x value where f(1) = 1 our graph point will lie on the graph of g.
When f = -1 we have g / f = -g, so at the corresponding x value our graph point will lie at the same distance from the x axis as the graph of g, but on the other side of the x axis.
When g = 1 we have g / f = 1 / f. (Note that if f is big, 1 / f is near 0; if f is near 0, then 1 / f is big; if f is positive, 1/f is positive and if f is negative 1/f is negative; if | f | > 1 then | 1/f | < 1 and if | f | < 1 then | 1 / f | > 1 ). In this case, when g = 1 we see that f > 1 but not real big so 1 / f will be < 1 but not real small. f is positive so 1/f will be positive. A specific estimate for this graph is that f = 4, approx., so 1/f = 1/4, approx.. We locate this point on the graph.
Where g = -1 we see that f > 1, and f is pretty big. So g / f = -1 / f will be negative and pretty close to 0.
Between the f = 1 point and the f = 0 point we can say that f is positive but < 1. g stays positive on this interval so g / f will remain positive. Since | f | < 1, we can also say that | g / f | > g.