Example 3 (Continued): Sketching a Parabola

EXAMPLE 3 (Continued)

Sketch the parabola y = 2x² - 5x + 4. The first two derivatives are

The only critical point is at the vertex, where x = 5/4. The table of values follows.

The parabola is drawn in Figure 5.4.10, again using Steps 1-5.

Figure 5.4.10

We can now sketch the graph of any equation of the form

Ax² + Dx + Ey + F = 0.

In the ordinary case where both A and E are different from zero, proceed as follows. First, solve the equation for y, obtaining the new equation

Second, use the method in this section to sketch the graph, which will be a vertical parabola. There are also two degenerate cases. If A = 0, the graph is a straight line. If E = 0, then y does not appear at all, and the graph is either two vertical lines, one vertical line, or empty.

We can also sketch the graph of any equation of the form

Cy² + Dx + Ey + F = 0.

In the ordinary case where C and D are different from zero, the graph will be a horizontal parabola.