Any polynomial function is said to be an expression of the form p(x) = a0 + a1*x + a2 * x2 + a3* x3 . . . . + an * xn , where we have an< > 0, is called the polynomial Functions, ‘x’ in the degree of ‘n’. While exploring polynomial Functions, we say that a0, a1, a2, a3… are all the real Numbers and each power of ‘n’ is a non negative Integer.
Exploring Quotients Polynomial FunctionsBack to Top
1. Horizontal asymptotes
2. Vertical asymptotes
3. Oblique asymptotes
To understand exploring graphs Polynomial Functions let us consider some important cases.
Case 1: Degree of polynomial expression present in numerator is smaller as compared to the denominator because denominator is growing faster than numerator, horizontal Asymptote remains 0 always. An example of it can be:
Y = X / X2 – 1
Or Y = X / (X – 1) (X + 1)
Here we see that graph of above function approaches infinity at X = 1 and X = -1 (Vertical Asymptotes). Thus these values are to be removed from Domain of function.
Case 2: This case discusses about same degree of polynomials in both numerator and denominator. Here also we get horizontal asymptotes which are given as Ratio of coefficient of highest degree term in polynomials i.e. coefficients in numerator to coefficient in denominator. An example of it can be:
Y = 2x / x – 1,
Here asymptote will be found at X = 2 / 1 (Horizontal Asymptote).
Case 3: Degree of polynomial in numerator is one greater than denominator. In this case an oblique asymptote is also generated.
Suppose we have a function y = x2 + 2x + 1 / x – 1,
In this example we get an oblique asymptote y = x + 3 and a Vertical Asymptote at x = 1.
Inverse Functions and RelationsBack to Top
f -1 (x) = (s, t): (4, 2), (1, 5), (3, 9), (7, 3).
Relation: Let S and T are two Sets then relation R defined from set S to set T, is a subset of S * T. Relation is represented in list form and in tabular form as well, for example a relation ‘R’ on set S = 1, 2, 3, 4, 5 defined by R = (S, T): T = S + 2 can also be expressed as:- S R T if and only if T = S + 2. Now we will see the inverse of function and relation.
First we will talk about Inverse Function: Suppose we have a function ‘P’ then its inverse form is denoted by P-1, we can say that relation can be obtained by interchanging each variable of function (p, q) of f by (q, p) of function f-1. For example: If we have a function f (6, -9) then inverse of given function is f-1 (-9, 6).
In the case of inverse relation: First we have to find the inverse relation for given function. Let given function f (p) = 3p + 2q = 3p + 2. After interchanging the value of P and Q it can be written as: 3q + 2 = p, if we want to find the inverse of function then subtract the function by 2 and then divide it by 3. Now we need to solve the function for y. In this way we can find the inverse of function and relation.
Remainder and Factor TheoremsBack to Top
A (y) / B (y) = Q (y) with remainder R (y). This can also be written as
A (y) = B (y) * Q (y) + R (y) where A (y) is dividend, B (y) is divisor, Q (y) is quotient and R (y) is remainder. Factors are used to determine the solution of given quadratic or polynomial equation. Factors of a given equation can be done using Quadratic Formula. There are two theorems, remainder and factor theorem, used for solution of division and Factorization. In remainder theorem, when we divide a polynomial f (y) by y - k, we get, f (y) = (y – k) * q (y) + r (y). But since r (y) is a constant term that’s why we use only 'r' in place of r (y), this gives f (y) = (y – k) * q (y) + r,
When y = k then,
f (k) = (k – k) * q (k) + r,
f (k) = (0) * q (k) + r,
f (k) = r,
Thus according to remainder theorem, when we divide a polynomial f (y) by y – k, then remainder 'r' will be f (k).
Remainder and factor theorems are related to each other. According to factor theorem if f (k) equals to 0 then remainder is also 0 and (y – k) is the factor of the polynomial.
Let’s consider the following example
F (k) = y2 – 3y - 4 = 0, in this equation when we put k = 4 gives, F (4) = (4)2- 3 (4) – 4 = 16 – 12 - 4 = 0 and hence (y - 4) must be factor of y2 – 3y - 4. Thus according to Factor Theorem, if f (k) = 0 then y – k is factor of the polynomial.
Roots and ZerosBack to Top
P (1) = 5 (1)3 - 4 (1)2 + 7 (1) – 8 = 5 – 4 + 7 – 8 = 0.
P (1) = 0 at y = 1, hence y = 1 is a root of polynomial equation.
If x + y j is zero (root) then x – y j is also a zero of polynomial function.
Let’s consider an example to show that if (2 + j) is a zero of f (y) = - y 2 + 4 y – 5, then (2 – j) is also a zero of the function.
First we will check if (2 + j) is a zero to f (y) by putting zero in given function:
f (y) = - y 2 + 4 y – 5,
= - (2 + j) 2 + 4 ( 2 + j) – 5,
= - (4 + j2 + 4j) + 8 + 4j – 5,
= - (4 – 1 + 4j) + 3 + 4j,
= - 3 – 4j + 3 + 4j = 0
f (y) = 0.
(2 + j) is a zero, so (2 – j) also must be a zero; plug (2 – j) in the function:
f (y) = - y 2 + 4 y – 5,
= - (2 - j) 2 + 4 (2 - j) – 5,
= - (4 + j2 - 4 j) + 8 - 4 j – 5,
= - (4 – 1 - 4 j) + 3 - 4 j,
= - 3 + 4 j + 3 - 4 j = 0, thus
f (y) = 0.
Rational Zero TheoremBack to Top
Rational zero theorem is defined as a polynomial function which is written in order of exponents which are descending in nature then, form of rational zero will be of the order ±p/ q. Here, p and q are defined as factors of constant term and factor of leading coefficient respectively. Let us consider a function as an example:
f(y) = 2y3 + 3y2 – 8y +3. According to rational zeros theorem, rational zero of numerator must be having a factor of 3 in its numerator part and in the denominator part also it must be having a factor of 2. Thus, for numerator or for ‘p’ factors of 3 that is equals to ±1, ±3 and for denominator or for ‘q’ factors of 2 are ±1, ±2. And according to rational zeroes theorem p/ q form in its simplest form is ±1, ±1/2, ±3, ±3/2. The values of zeroes can be found by just using substitution or we can say direct substitution or using division and finding the remainder.
Polynomial FunctionsBack to Top
f (y) = b m y m + b (m – 1) y (m – 1) +…+ b 2 y 2+ b 1 y + b 0
Here 'm' is non negative Integer and b 0, b 1… b (m – 1), b m are constant coefficients. Polynomial Functions can be with one or more than one terms. For instance, function ƒ(y) is defined as
f (y) = y 3 – y is a polynomial function with one argument. Polynomial functions of multiple arguments can be defined as:
f (x, y) = 3x 3 + 4 x 2 y + 5 x y 5 + 8
Term which has a variable with highest exponent is called as leading term. In polynomial functions, leading term is b m y m . In above example, it is 3x 3 is leading term. Leading coefficient can be defined as coefficient of term with highest exponent. Leading coefficient in polynomial function is (b m).
Degree of terms of polynomial function can be defined as power (exponent) on variable. Degree of standard form of polynomial function is 'm'.
Let's consider an example to understand the concept of polynomial function. Consider
f (y) = -4y5 + 33y4 + 3y2 + 12.
In this example, since a variable with highest exponent is called as leading term hence leading term is -4y5. Highest exponent is 5 hence degree of given polynomial function is 5. Leading coefficient is the multiplying factor of variable with highest exponent (power) hence leading coefficient for above equation is -4. Polynomials are of different types like monomial, binomial, trinomial etc.