Algebra Know-It-ALL

we must know the coordinates of another point. We can find one by moving horizontally to the right
by any number of units we want (call it n units), and then moving straight up from there by mn units,
where m is the slope. We should move far enough to the right so the two points will be well separated.
That will make it easy to draw the graph accurately. Let’s move to the right from (0, 3) by 3 units. That
gives us the point (3, 3), shown as an open circle to indicate that it’s not actually part of the graph. Then
we move straight up by mn=− 2 × 3 =−6 units, which is equivalent to moving straight down by 6 units.
We have now found a second point on the graph. Our new point is 3 units to the right and 6 units below
they-intercept point, so its coordinates are (3, −3). We plot this point, and then we connect the two
points with a solid line to obtain the graph. We extend the line somewhat beyond the points in either
direction, keeping in mind that the true, complete line (in the mathematical cosmos) extends infinitely
far in each direction!

Here’s another challenge!

Just as there is a standard form for a first-degree equation in one variable, there’s a standard form for a
two-variable linear equation. Here it is:

ax+by+c= 0

where x is the independent variable, y is the dependent variable, and a,b, and c are constants. Show how
this equation can be rearranged into the SI form, expressing y as a function of x (as long as b, the coefficient
ofy, is not equal to 0).

–6 4 6

2

6

–2

–4

–6

x

y

–4 –2

Move to right

by 3 units

Move up

by –6 units

which is

down by

6 units

y-intercept

is 3

Slope = –2

Figure 15-4 Graph of the equation 8x+ 4 y= 12. This

graph can be drawn easily when we morph

the equation into its SI form y=− 2 x+ 3.

Slope-Intercept Form 241