Algebra Know-It-ALL

dard PS equation, we get

y − 2 = 2[x − (−1)]

which simplifies to

y − 2 = 2(x+ 1)

We can now find another point on the graph by plugging in a value for x and solving for y. We
only need one more point to determine the straight line in the Cartesian plane that represents
this function. Let’s try x= 1. We solve for y in steps:

y − 2 = 2(1 + 1)

y − 2 = 2 × 2

y − 2 = 4

y= 6

This tells us that (1, 6) represents a point on the graph. We already know that (−1, 2) is on it.
When we plot these two points on the plane and draw a straight line through them both, we
get the graph shown in Fig. 15-5.

–6 246

4

6

–2

–4

–6

x

y

–4 –2

(x 0 ,y 0 ) = (–1,2)

(1,6)

m= 2

Figure 15-5 Graph of a linear function based on the

knowledge that the point (−1, 2) is on the

line, and the slope is equal to 2.

Point-Slope Form 243