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482
ALGEBRA
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482

482 ALGEBRA.

Ex. 1. Construct the graph of 2x —1. Let 2x—1=y. Giving to x successive values, we obtain the corresponding values of y as follows: x=—2, y=—5. x=—1, y=—3: x =0, y= —1 x=1, y=1 x = 2, y = 3 x= 3, y=5

Locating these points as explained in Art. 611 and drawing a line through them, we have in this case the straight line AB, Fig. 2, as the graph required.

Ex. 2. Plot the equation x^2 — 2x-4.

Putting y= x^2-2x-4, we obtain the following values: x=—2, y=4. x=-1, y=-1 x=0, y=-4. x=1, y=-5. x=2, y=—4. x=3, y= -1. x=4, y=4.

The points lie on the line ABC, Fig. 3, which is the graph of the given equation.

Ex. 3. Plot the graph of x^3-2x.

Assuming y = x^3-2x, we have

e=-2, y=-4. x= -1, y = -1. x=0 , y = 0. x=1 , y = 1. x=2 , y = 4.

The line ABCD, Fig. 4, is the required graph.

By taking other values between those assumed, we may locate the curve with greater precision.