Distance, Circles, And Quadratic Equations Worksheets Page 6
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9H6
Appendix H: Distance, Circles, and Quadratic Equations
With the aid of this formula, a reasonably accurate graph of a quadratic equation in x can
y = x
− 2x − 2
2
x
be obtained by plotting the vertex and two points on each side of it.
−1
1
−2
0
−3
1
Example 9
Sketch the graph of
−2
2
(a) y = x
− 2x − 2
(b) y = −x
+ 4x − 5
2
2
3
1
y
The equation is of form (7) with a = 1, b = −2, and c = −2, so by (8)
Solution (a).
the x-coordinate of the vertex is
b
x = −
= 1
2a
Using this value and two additional values on each side, we obtain Figure H.8.
x
The equation is of form (7) with a = −1, b = 4, and c = −5, so by (8)
Solution (b).
−2 −1
1
2
3
4
the x-coordinate of the vertex is
b
x = −
= 2
2a
Using this value and two additional values on each side, we obtain the table and graph in
Figure H.9.
y = x
− 2x − 2
2
Figure H.8
Quite often the intercepts of a parabola y = ax
+ bx + c are important to know.
2
The y-intercept, y = c, results immediately by setting x = 0. However, in order to obtain
the x-intercepts, if any, we must set y = 0 and then solve the resulting quadratic equation
+ bx + c = 0.
2
ax
y = −x
+ 4x − 5
2
x
−5
0
−2
Example 10
Solve the inequality
1
−1
2
− 2x − 2 > 0
2
x
−2
3
−5
4
Solution.
Because the left side of the inequality does not have readily discernible factors,
y
the test-point method illustrated in Example 4 of Appendix F is not convenient to use.
Instead, we will give a graphical solution. The given inequality is satisfied for those values
x
of x where the graph of y = x
− 2x − 2 is above the x-axis. From Figure H.8 those are
2
−1
1
2
3
4
5
the values of x to the left of the smaller intercept or to the right of the larger intercept. To
find these intercepts we set y = 0 to obtain
− 2x − 2 = 0
2
x
Solving by the quadratic formula gives
√
√
√
−b ±
− 4ac
2 ±
2
12
b
x =
=
= 1 ±
3
2a
2
y = −x
2
+ 4x − 5
Thus, the x-intercepts are
Figure H.9
√
√
x = 1 +
3 ≈ 2.7 and x = 1 −
3 ≈ −0.7
and the solution set of the inequality is
√
√
(− , 1 −
3) ∪ (1 +
3, + )
Note that the decimal approximations of the intercepts calculated in the preceding example agree
REMARK
with the graph in Figure H.8. Observe, however, that we used the exact values of the intercepts to
express the solution. The choice of exact versus approximate values is often a matter of judgment that
depends on the purpose for which the values are to be used. Numerical approximations often provide
a sense of size that exact values do not, but they can introduce severe errors if not used with care.
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