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PROBLEM 29, OCTOBER 2005Author(s): Robert O. StantonSource: The Mathematics Teacher, Vol. 100, No. 3 (OCTOBER 2006), pp. 172-173Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27972177 .
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ons
VERTICES AND x-INTERCEPTS OF PARABOLAS (I) I enjoyed reading James Metz's letter in the "Reader Reflections" of February 2006.1 would like to propose a general ization of his method of finding the = coordinate of the vertex of a parab
ola and use the result to find the zeroes of the parabola.
For/(x) = ax2 + bx + c, then is the
vertex of the parabola if/(x + d) =
f{x -
d) for any d.
Let's find x. Substitution leads to:
a{x + d)2 + b{x + d) + c = a{x- d)2 +
b{x -
d) + c ?> 4:axd = -2bd
As d = 0 is of no interest, we divide by d to get
This result leads to the quadratic for mula without completing the square. Let's see how.
It is easy to find the zeroes of the
parabola fix) = ax2 + c whose axis of
symmetry is the y axis.
Every parabola fix) = ax2 + bx + c can
be translated so that its axis of symmetry is the y axis:
is the translated parabola. Solving
will give us the two zeroes of the translated parabola. We shall find the zeroes:
2a
-+b\ -+c =
2a) { 2a)
a 2 - 2axb ab2
+ ?- + bx-+ c = 0 4a2 2a 2a
b' ,2 b' ,2 2 V r\ ax +-+ c = 0
4a 2a
2 b2 b2 ax =-c
2a 4#
2b2-b2-4ac b2-4ac
These are the zeroes of the translated
parabola, and it is easy to see the zeroes
of the original one.
Many students fail to see the con
nection between the solutions of the
quadratic equation and the zeroes of the
parabola. Proving the famous formula
again in the setting of functions can be
interesting and productive. This method also provides us with an application of translations.
Ayana Touval
Ay ana. Touval @ montgomery college, edu
Montgomery College Rockville, MD 20850-1733
x = ?
VERTICES AND x-INTERCEPTS OF PARABOLAS (II) In "Finding the x-Coordinate of the Vertex of a Parabola" ("Reader Reflec
tions," Mathematics Teacher 99 [Feb ruary 2006]: 390) James Metz gives another, simpler, noncalc?lus way of finding the vertex of a parabola. It should be noted that the vertex can be found more directly using calculus.
Consider the equation of a parabola, the function f[x)
= ax2 + bx + c. Its deriv
ative, the slope of the curve, is/(x) =
lax + b. The x-coordinate of the vertex is found from/(x0)
= 0, the position of zero slope. It follows immediately that
-b x=?. 0 2a
Of course, the ?/-coord?nate also follows
immediately:
-(b2-4:ac)
JohnF. Goehljr. jgoehl @ mail harry, edu
Barry University Miami Shores, FL 31161
PROBLEM 29, OCTOBER 2005 "Calendar" problem 29, from October
2005, reads:
Suppose 0 is an angle between 0? and 90? for which cos (0) cos (20) = 1/4.
What is the value of 0?
The solution presented is obtained by multiplying both sides of the equation by 4 sin 0 and using a trigonometric iden
tity to obtain sin (40) = sin (0). Relying on the fact that 0? < 0 < 90?, it is shown that 0 = 36?. The extraneous root 0 = 0?
must be rejected. In the spirit of one of the excellent
departments of the Mathematics Teacher, let us engage in "delving deeper." Specif ically, we will drop the restriction on , and find the general solution of cos (0) cos (20) = 1/4. Let be the set of inte
gers. By the periodic nature of the cosine
function, 0 is a solution precisely if 0 +
360?ft is a solution, for e . Since cos 0 is an even function, 0 is a solution if and only if -0 is one. Therefore, it suf fices to restrict the consideration to the interval 0 g [0?, 180?].
As in the presented solution, the restriction of 0 to the first quadrant forces 40 to be in the second quadrant, and 0 = 36?. When 0 g [90?, 180?], 40 g [360?, 720?]. However, for sin
(40) = sin (0) to be true, sin (40) must be positive, which reduces the interval to 40 g [360?, 540?], angles in the first and second quadrant. When 40 is in the first quadrant, it is 360? greater than reference angle 180? - 0:
40 = 360? + 180? - 0 50 = 540? 0=108?
If 40 is in the second quadrant,
40 = 360? + 0 30 = 360? 0=120?.
Therefore the general solution is
36?+ 360?^, 108? + 360? ?2, 120?+ 360? n3, -36?+360 , -108?+360 ,
-120? + 360? n6;
?l>?2'?3>n4?n5'ne G Z
172 MATHEMATICS TEACHER | Vol. 100, No. 3 ? October 2006
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An alternative, although more diffi
cult, approach is instructive. In the given equation cos (0) cos (20) = 1/4, replace cos (20)by2cos20-l:
cos (0) cos (20) = 1/4 cos (0)(2 cos20-l) = l/4 4 cos (0)(2 cos20-l) = l 8 cos30 - 4 cos 0 - 1 = 0
Letting = cos 0 yields the cubic equa tion 8z3 - 4z - 1 = 0. The rational root theorem finds the root = (-1/2), and the quadratic formula completes the solution:
= - l?y?5
Restricting 0 to [0?, 180?] as above, we
immediately find the root 0 = 120? cor
responding to = (-1/2). A computer algebra system shows that
arceos - = 36
and
arceos = 108?.
This approach produces no extraneous roots.
It is instructive to view the graphs of the functions in question: y = cos
(x) cos (2x), in blue, and y = 1/4, in
red, are shown on the interval [-180?, 180?] in figure 1 (Stanton). The six roots in this interval are represented by the intersections of the two graphs. Note the symmetry with respect to the
?/-axis.
Figure 2 (Stanton) shows y = sin
(x) in red and y = sin (4x) in blue. Aside from the extraneous roots when y = 0, the intersections of the two graphs cor
respond to the solutions of the original equation. As expected with sine func
tions, the graphs display symmetry about the origin.
Robert O. Stanton
[email protected] St. John's University
Jamaica, NY 11439 <?
Fig. 1 (Stanton)
Fig. 2 (Stanton)
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