Suppose that our quadratic expression Q(x) = ax2 + bx + c is written in1
factored form a(x − p)(x − q) where p < q.
We describe how to make a similar analysis depending on the sign of a. First, we need a good notation to report our answers efficiently.
1In general this cannot be done with p and q real. However solving hard mathematical problems
almost always turns on finding the right special cases to study for insight. It is a maxim among mathematicians that “If there is a hard problem you can’t solve then there is an easy problem you can’t solve”.
6.2. FACTORED QUADRATIC POLYNOMIAL. 87
Interval notation.
Given real numbers a < b we shall use the following standard convention. 2
Values of x Notation
a < x and x < b (a, b) a ≤ x and x < b [a, b) a < x and x ≤ b (a, b] a ≤ x and x ≤ b [a, b]
A simple way to remember these is to note that we use open parenthesis when the end point is not included, but use a closed bracket if it is!
The notation is often extended to allow a to be −∞ or b to be ∞.
Thus (−∞, 4) denotes all real numbers x such that −∞ < x < 4 and it clearly has the same meaning as x < 4, since the first inequality is automatically true!
As an extreme example, the interval (−∞, ∞) describes the set ℜ of all real numbers!
The reader should verify the following: 1. Case a > 0 Remember that
Q(x) = a(x − p)(x − q) with p < q. Then we have: Values of x behavior of Q(x) (−∞, p) Q(x) > 0 x = p Q(x) = 0 (p, q) Q(x) < 0 x = q Q(x) = 0 (q, ∞) Q(x) > 0
See the graphic display below showing the results when a > 0.
Zero at q Zero at p
Positive Negative
Positive
How do you verify something like this? Here is a sample verification. Consider the third case of the interval (p, q). Since p < x < q in this interval, we note that:
x − p > 0 and q − x > 0 which means the same as x − q < 0.
2The a and b used in this definition have nothing to do with our coefficients of the quadratic!
Since Q(x) = (a)(x − p)(x − q) and since a > 0, we see that Q(x) is a product of two positive terms a and (x − p) and one negative term (x − q).
Hence it must be negative!
In general, if our expression is a product of several pieces and we can figure out the sign for each piece, then we know the sign for the product!
2. Case a < 0
Everything is as in the above case, except the signs of Q(x) are reversed! To simplify our discussion, let us make a few formal definitions.
Definition: Absolute Maximum value Given an expression F (x) we say that it has an absolute maximum at x = t if F (t) ≥ F (x) for every x ∈ ℜ. We will say that F (t) is the absolute maximum value of F (x) and that it occurs at x = t.
We may alternatively say that x = t is an absolute maximum for F (x) and the absolute maximum value of F (x) is F (t).
We similarly make a Definition: Absolute Minimum value Given an expres- sion F (x) we say that it has an absolute minimum at x = t if F (t) ≤ F (x) for every x ∈ ℜ. We will say that F (t) is the absolute minimum value of F (x) and that it occurs at x = t.
We may alternatively say that x = t is an absolute minimum for F (x) and the absolute minimum value of F (x) is F (t).
Definition: Absolute Extremum value A value x = t is said to be an absolute extremum for F (x), if F (x) has either an absolute maximum or an absolute minimum at x = t.
Set
m = (p + q)
2 and u = x − m.
Note that m is the average value of p and q and also x = u+m. Note the following algebraic manipulation:
Q(x) = a(u + p+q2 − p)(u +p+q2 − q) = a(u − p−q2 )(u + p−q2 )
= a(u2− T2) where T = p−q 2
Since T2 is a positive number and u2 is always greater than or equal to zero, the
smallest value that u2− T2 can possibly have is when u = 0. However this happens
exactly when x = m. This shows that u2 − T2 = (x − m)2 − T2 has an absolute
minimum value at x = m. If we take x large enough then u2 can be made as large as
we want. This means that there is no upper bound to the values of u2− T2
It follows that if Q(x) that can be put in factored form Q = (a)(x − p)(x − q) with positive a, then Q(x) has an absolute minimum at x = m, where m is the average of p, q.
6.2. FACTORED QUADRATIC POLYNOMIAL. 89
In case Q(x) = a(x − p)(x − q) with a negative a, the situation reverses and Q(x) has an absolute maximum (and no absolute minimum) at x = m, the average of p, q. Thus, in either case, we can say that Q(x) has an extremum at x = m the average of p, q.
We shall show below that indeed, this value x = m is actually x = −b/(2a). Thus the important point is that we need not find the values p, q to find this extremum value!
Remark. Note that for a general expression, the absolute maximum or minimum value may not exist and may occur at several values of x.
Here are some examples to clarify these ideas.
1. As already observed, a linear expression like 3x − 5 does not have a absolute maximum or minimum; its values can be arbitrarily small or large!
2. Consider the expression
F (x) = (x2− 1)(2x2− 7).
We invite the reader to use a graphing calculator or a computer to sketch a graph and observe that it has value −25/8 = −3.125 at x = −1.5 as well as x = 1.5.
It is easy to see from the graph that it has an absolute minimum value at these values of x. It is also clear that there are no absolute maximum values.
The complete argument to prove this without relying on the picture requires a further development of calculus and has to be postponed to a higher level course.
Abs. Min. Abs. Min.