From the quadratic formula, we know that the sum of roots of even this equation is always −ab and hence the average of the roots is again −
b 2a.
Thus, if we were to plot the graph of y = Q(x) − k, we get a parabola and its extremum value occurs at the x-coordinate of the vertex of the parabola, namely x = −2ab .
Thus, the extremum value of the function Q(x) − k is the y-coordinate of the vertex of the parabola y = Q(x) − k. Thus:
The extremum value of Q(x) − k occurs at x = −2ab and is equal to Q(−2ab )−k.
This is sometimes called the Vertex Theorem. .
6.4
Examples of Quadratic polynomials.
We discuss a few exercises based on the above analysis.
1. Find the extremum values for Q(x) = 2x2+ 4x − 6 and determine interval(s) on which Q(x) is negative.
Answer: We have a = 2, b = 4, c = −6. Thus − b2a = −44 = −1. So, if we replace x = u − b2a = u − 1, then we have:
Q(x) = 2(u − 1)2+ 4(u − 1) − 6
= 2u2− 4u + 2 + 4u − 4 − 6
= 2u2− 8 = 2(u2− 4)
We have shown the simplification by hand, but if you remember the theory, you could just write down the last line from Q(x) = a(u2− H) thus:
a = 2 and H = b 2− 4ac 4a2 = 16 − 4(2)(−6) 4(22) = 64 16 = 4.
At any rate, the value of Q(x) is clearly bigger than or equal to 2(−4) = −8 and this
absolute minimum −8 is attained when u = 0 or x = u − 1 = −1. Finally, the quadratic formula (or straight factorization) will yield two roots p = −3, q = 1.
114CHAPTER 6. SPECIAL STUDY OF LINEAR AND QUADRATIC POLYNOMIALS.
If you plot these two values on the number line you get three intervals. To check the sign of Q(x), the easiest way is to test a few strategic points as we did in the linear case.
Positive Negative Positive B(1) A(–3) Intervals (−∞, −3) −3 (−3, 1) 1 (1, ∞) Test points −4 −3 0 1 4 Values 10 0 −6 0 42
Conclusion Positive Zero Negative Zero Positive Thus, the expression is negative exactly on the interval (−3, 1). 2. Find the extremum values for Q(x) = 2x2+ 4x + 6 and determine interval(s)
on which Q(x) is negative.
Answer: Now we have a = 2, b = 4, c = 6, so
H = b
2− 4ac
4a2 =
16 − 4(2)(6)
4(4) = −2.
This is negative, so the theory tells us that Q(x) will have its extremum at x = − b2a = −1 and this extremum value of Q(x) is 2(−1)2+ 4(−1) + 6 = 4.
The theory says that the expression will always keep the same positive sign and so 4 is actually the minimum value and there is no maximum.
Chapter 7
Functions
7.1
Plane algebraic curves
The simplest notion of a plane algebraic curve is the set of all points satisfying a given polynomial equation f (x, y) = 0.
We have seen that a parametric form of the description of a curve is useful for generating lots of points of the curve and to understand its nature.
We have already learned that all points of a line can be alternately described by linear parameterization x = a + ut, y = b + vt for some suitable constants a, b, u, v where at least one of u, v is non zero.
In case of a circle (or, more generally a conic) we will find such a parameterization, but with rational functions of the parameter t, rather than polynomial functions.
Definition: A rational Curve. We say that a plane algebraic curve f (x, y) = 0 is rational if it can be parameterized by rational functions.
This means that we can find two rational functions x = u(t), y = v(t) such that at least one of u(t), v(t) is not a constant and f (u(t), v(t)) is identically zero.
Actually, to be precise, we can only demand that f (u(t), v(t)) is zero for all values of t for which u(t), v(t) are both defined.
Examples. Here are some examples of rational algebraic curves. You are advised to verify the definition as needed.
1. A line. The line y = 3x + 5 can be parameterized as x = t − 2, y = 3t − 1. Check:
(3t − 1) = 3(t − 2) + 5.
2. A Parabola. The parabola y = 4x2+ 1 can be parameterized by x = t, y = 4t2+ 1.
3. A Circle. The circle x2 + y2= 1 can be parameterized by
x = 1 − t 2 1 + t2, y = 2t 1 + t2. 115
116 CHAPTER 7. FUNCTIONS
You need to verify the identity:
(1 − t 2 1 + t2) 2+ ( 2t 1 + t2) 2 = 1
for all values of t for which 1 + t2 6= 0. Since we are working with real numbers
as values, this includes all real numbers.
4. A Singular Curve. The curve y2= x3 is easily parameterized by
x = t2, y = t3. Note that a very similar curve y2 = x3− 1 is declared not to be
parameterizable below.
5. A Complicated Curve. Verify that the curve: xy3− x2y2+ 2 xy = 1 can be parameterized by:
x = t
3
t2− 1, y =
t2− 1
t .
This is simply an exercise in substitution. The process of deciding when a curve can be parameterized and then the finding of the parameterization, both belong to higher level courses.
It is tempting to think that all plane algebraic curves could be parameterized by rational functions, but alas, this is not the case. The simplest example of a curve which cannot be parameterized by rational functions is the curve given by
y2 = x3− 1.1
Since we cannot hope to have all algebraic curves parameterized, we do the next best thing; restrict our attention to the ones which are! In fact, to get a more useful theory, we need to generalize the idea of an acceptable function to be used for the parameterization. We may see some instances of this later.
1The proof that a given curve is not rational is not elementary and requires new ideas, found
in advanced books on Algebraic Geometry. Pursuing this example is a good entry point into this important branch of mathematics.