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Example 3 (Continued): Sketching a Parabola

EXAMPLE 3 (Continued)

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

05_limits_g_approx-270.gif

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

x

y

dy/dx

d2y/dx2

Comments

limx→-∞

-∞

vertical

5/4

7/8

0

+

min, ∪

limx→∞

vertical

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

05_limits_g_approx-272.gif

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

05_limits_g_approx-271.gif

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.


Last Update: 2006-11-05