Algebra Know-It-ALL

Part Two 321

Answer 14-9

Using the horizontal-line test, we can see that G and H represent functions of y, at least within

the viewing region. They both pass the test. But neither E nor F represents functions of y,

because they fail the horizontal-line test. (Part of curve E is itself a horizontal line.)

Question 14-10

Can we restrict the domains of the inverses of relations E or F to make either of them into a

nontrivial function of y? If so, how?

Answer 14-10

In the case of the inverse of the “bent line” curve E, we can make it into a function if we

confine the domain to values of y strictly larger than 1. This is based on the assumption that

the lines that make up curve E continue straight for infinite distances in both directions. The

inverse of curve F is not so cooperative. Imagine “taking slices” of by considering only that

portion of the curve that falls between two movable horizontal lines. This limits the values of y

that apply to the curve, thereby restricting the domain. No matter how we “slice it,” we always

get two values of x for each value of y. The only way we can get a function out of the inverse

of curve F is to bring the two horizontal lines together so they both intersect the graph at the

extreme top point or the extreme bottom point. Those are trivial results.

Chapter 15

Question 15-1

Figure 20-5 is a Cartesian graph showing the same three points P,Q, and R as we saw

in Fig. 20-3. Three lines, each extending indefinitely in either direction, pass through pairs

of these points. Let’s call the lines PQ,QR, and PR. (It’s visually apparent which is which!)

Assume that the ordered pairs for the points are all pairs of integers. In other words, assume

that the points are exactly where they appear to be. Based on this information, how can we

determine the slope of line PQ? How can we determine the y-intercept of line PQ?

Answer 15-1

The slope of a line is equal to the change in the y-value (Δy) divided by the change in the

x-value (Δx) as we move from one point on the line to another point. We know the ordered

pairs for the two points as P= (−5,−3) and Q= (0,−1). Therefore,

Δy=− 1 − (−3)

=− 1 + 3

= 2

and

Δx= 0 − (−5)

= 0 + 5

= 5