# The A to Z of mathematics a basic guide - Sidebotham T.H.

ISBN 0-471-15045-2

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x =--------------------

2 a

The example that follows demonstrates how to solve a quadratic equation using this formula.

Example. Solve the equation 2x2 — 3.x — 6 = 0 to 2 dp.

Solution. Comparing 2.x2 — 3x — 6 = 0 with ax2 + bx + c = 0, we have a = 2, b = —3, c = —6. Write

Substituting for a, b, c into —b ± V b2 — 4ac

Separating out into two answers

dp)

Reference: Completing the Square, Quadratic Equations.

QUADRATIC GRAPHS

The graph of the quadratic expression y = ax2 + bx + c, where b or c, or both may be zero, is a called a parabola. In this formula a f 0, and a, b, and c are real numbers. The parabolas we study may be “cup”-shaped, in which case a > 0, or “hat”-shaped, in which case a < 0 (see figure a).

■(-3) ± V(-3)2 - 4 x 2 x (-6) 2x2

3± V9T48 4

3 ± V57

3 + V57 3-V57

X = --------- or X = -------

x = 2.64

or x = — 1.14 (to 2

QUADRATIC GRAPHS 363

a > 0 a < 0

(a)

The most basic quadratic graph is y = x2, and its graph is drawn in figure b. The points used to draw this graph are

(-3, 9), (—2, 4), (-1, 1), (0, OX (1, IX (2, 4), (3,9)

In fact all the graphs of y = x2 + bx + c, that is, when a = 1, are exactly the same as y = x2, except they have been given a translation so that the vertex of each graph is at a different point. The theory of the translation of parabolas is explained here.

The equation of the parabolic graph y = x2 + bx + c can be expressed as y = (x + h)2 + v, where h and v are constants, using a process called completing the square. When the equation is expressed in this form it is easier to sketch its graph. The rales for doing this are explained in the examples that follow.

Example 1. Sketch the graph of y = (x — 2)2.

Solution. By comparing y = (x — 2)2 with y = (x + h)2 + v, we see that h = — 2 and v = 0. This value of h means the graph of y = x2 is given a horizontal translation of two squares to the right to obtain the graph of y = (x — 2)2. The result of this process is shown in figure c with y = x2 drawn dashed; y = (x — 2)2 has its vertex at the point (2, 0).

364 QUADRATIC GRAPHS

The graph of y = (x + 2)2 would have its vertex at the point (—2, 0). This is because h = +2 and the graph of y = x2 is translated two squares to the left. The graph of y = x2 can also be translated vertically, as shown in the next example.

Example 2. Sketch the graph of y — x2 — 3.

Solution. By comparing y = x2 — 3 with y = (x + h)2 + v, we see that h = 0 and v = —3. This value of v means the graph of y = x2 is given a vertical translation of three squares downward to obtain the graph of y = x2 — 3. The result of this process is shown in figure d with y = x2 drawn dotted; y = x2 — 3 has its vertex at the point

The graph of y = x2 + 3 would have its vertex at the point (0, 3). This is because v = 3 and the graph of y = x2 is translated three squares upward.

Summary. When graphing the equation y = (x + h)2 + v, where one unit on the graph is one square:

♦ If h is positive and v = 0, move y = x2 horizontally to the left through h squares.

♦ If h is negative and v = 0, move y = x2 horizontally to the right through h squares.

♦ If v is positive and h = 0, move y = x2 vertically upward through v squares.

♦ If v is negative and h = 0, move y = x2 vertically downward through v squares.

♦ When neither h nor v is zero, we sketch a combination of horizontal and vertical translations.

The same processes are applied to sketching the graphs of parabolas with the equation y = — (x + h)2 + v, except the graph is “hat”-shaped instead of “cup”-shaped.

Stretches of y = x2 So far we have taken the value of a in the equation of y = ax2 + bx + c to be either 1 or —1. For other values for a, we stretch the y = x2 in the horizontal direction and make it “flatter” or stretch y = x2 in a vertical direction and make it “steeper.” For example, the graphs in figure e are y = x2, which is shown dotted, y = 2x2, which is steeper, and y = | x2, which is flatter.

(0, -3).

0

x

-3

QUADRATIC GRAPHS 365

(e)

When the equation of parabolas are given in factorized form we can use the “intercept” method of sketching them. The method is simply to find the points where the parabolas cross the x- and y-axes and then sketch a “hat”- or “cup”-shaped parabola. This method produces a sketch graph that passes through the intercept points, and is explained using the examples that follow.

Example 3. Sketch the graph of the parabola y = (x — 3)(x + 1) clearly showing the curve’s axis of symmetry and vertex.

Solution. The equation of the curve after expanding the brackets is y = x2 — 2x — 3. Now that the brackets are expanded we can identify whether a is positive or negative. In this case the value of a is 1, which is positive, so the parabola is “cup”-shaped. The x intercepts are the points where the curve crosses the s-axis. To obtain these points we substitute y = 0 into the equation of the curve and solve the resulting equation:

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