Análisis beneficio-costo

When the argument changes in a linear function, the value of the dependent variable changes in constant proportion. That proportion can be positive or negative. It can even be zero, in which case we have a constant function.

Slope and intercept

In conventional coordinates, linear functions always produce straight-line graphs. Conversely, any straight line represents a linear function, as long as that line isn’t parallel to the dependent- variable axis.

Theslope, also called the gradient, of a straight line in rectangular coordinates (where the axes are perpendicular to each other and the divisions on each axis are of uniform size) is an expres- sion of the steepness with which the line goes upward or downward as we move to the right. A horizontal line, representing a constant function, has a slope of zero. A line that ramps upward as we move to the right has positive slope. A line that ramps downward as we move to the right has negative slope. Figure 1-4 shows a line with positive slope and another line with negative slope.

To calculate the slope of a line, we must know the coordinates of two points on that line. If we call the independent variable x and the dependent variable y, then the slope of a line, passing through two points, is equal to the difference in the y-values divided by the difference in the x-values. We abbreviate “the difference in” by writing the uppercase Greek letter delta (Δ). Let’s use a to symbolize the slope. Then

a= Δy/Δx 2 4 6 –6 2 4 6 –2 –4 –6 x y –4 –2 y-intercept is 3 y-intercept is –2 Slope is negative Slope is positive

Figure 1-4 Graphs of two linear functions.

We read this as “delta y over delta x.” Sometimes the slope of a straight line is called rise over run. This makes sense as long as the independent variable is on the horizontal axis, the depen- dent variable is on the vertical axis, and we move to the right.

Anintercept is a point where a graph crosses an axis. We can plug 0 into a linear equation for one of the variables, and solve for the other variable to get its intercept. In a linear func- tion, the term y-intercept refers to the value of the dependent variable y at the point where the line crosses the y axis. In Fig. 1-4, the line with positive slope has a y-intercept of 3, and the line with negative slope has a y-intercept of −2.

Standard form for a linear function

If we call the dependent variable x, then the standard form for a linear function is f (x)=ax+b

where a and b are real-number constants, and f is the name of the function. As things work out,a is the slope of the function’s straight-line graph. If we call the dependent variable y, then b is the y-intercept. We can substitute y in the equation for f (x), writing

y=ax+b

Either of these two forms is okay, as long as we keep track of which variable is independent and which one is dependent!

Are you confused?

If the graph of a linear relation is a vertical line, then the slope is undefined, and the relation is not a function. The graph of a linear function can never be parallel to the dependent-variable axis (or perpendicular to the independent-variable axis). In that case, the graph fails the verti- cal-line test.

Here’s a challenge!

Rewrite the following equation as a linear function of x, and graph it on that basis:

12x+ 6y= 18

Solution

We must rearrange this equation to get y all by itself on the left side of the equals sign, and an expres- sion containing only x and one or more constants on the right side. Subtracting 12x from both sides gives us:

6y= −12x+ 18

Dividing each side by 6 puts it into the standard form for a linear function: y= (−12x)/6+ 18/6 = −2x+ 3

If we name the function f, then we can express the function as f (x)= −2x+ 3

In the graph of this function, the y-intercept is 3. We plot the y-intercept on the y axis at the mark for 3 units, as shown in Fig. 1-5. That gives us the point (0,3). To find the line, we must know the coordi- nates of one other point. Let’s find the x-intercept! To do that, we can plug in 0 for y to get

0= −2x+ 3

Adding 2x to each side and then dividing through by 2 tells us that x= 3/2. Therefore, the point (3/2,0) lies on the line. Now that we know (0,3) and (3/2,0) are both on the line, we can draw the line through them.

Here’s a twist!

When we move from (0,3) to (3/2,0) in Fig. 1-5, we travel in the negative y direction by 3 units, so

Δy= −3. We also move in the positive x direction by 3/2 units, so Δx= 3/2. Therefore

Δy/Δx= −3/(3/2) = −2

reflecting the fact that the slope of the line is −2. We’ll always get this same value for the slope, no matter which two points on the line we choose. Uniformity of slope is characteristic of all linear functions. But there are functions for which it isn’t so simple.

4 6 –6 2 6 –2 –4 –6 x y –4 –2 Slope = –2 (0,3) y-intercept is 3 (3/2,0) x-intercept is 3/2 f(x) = –2x+ 3 y= –2x+ 3

Figure 1-5 Graph of the linear function y= −2x+ 3.