College Algebra Acosta/Karowowski

Polynomial and rational functions Chapter 7

Polynomial functions Chapter 7 – Section 1

Review and definition f(x) = x g(x) = x2 k(x) = x3 m(x) = x4 etc. Called power functions – a polynomial function is a sum of power functions where the exponents are whole numbers (ie – no fractions – no negative in exponents)

Which of the following are polynomial f(x) = 3x3 – 5x + 8 𝑔 𝑥 = 3𝑥 −5 +4 𝑥 3 L(x) = 3x – 7 𝑘 𝑥 = 6 𝑥 3 − 2 3 𝑥 7 𝑚 𝑥 = 6𝑥 +5𝑥

Other forms of polynomial functions Multiplying, adding, or subtracting polynomials always results in a polynomial Therefore polynomial functions may appear factored or as sums g(x) = (x – 9)3(x + 2)2 m(x) = (3x + 5)4 – (2x2 + 3)3

Degree of a polynomial/leading coeffiecient When the polynomial is simplified and written in descending order the degree of the polynomial is the degree of the leading term (in one variable polynomials this is simply the exponent of the leading term) Ex: g(x) = 3x5 – 2x3 + 18x2 + 10 It is neither necessary nor desirable to simplify a factored polynomial in order to determine its degree f(x) = (x – 2)2(x – 5)3(x + 2) k(x) = (4x2 – 7)2 (x3 + 6) m(x) = (3x + 5)4 – (2x2 + 3)3

Analyzing polynomial functions- what the equation (or the graph) tells us Domain Range Maximums, minimums - end behavior y- intercept x – intercept Turning points – local/relative maximums and minimums General shape – end behavior (limits)

Domain There are no restrictions on power functions, multiplication, addition or subtraction so there are NO restrictions on polynomial functions Conclusion - domain for ALL polynomial functions is - all real numbers

Range/ max -min Since the domain is unrestricted the question of range translates into “Is there a maximum or minimum value?” With power functions – even powers attain a maximum or minimum while odd powers do not What happens when we add/subtract even and odd power?

Conclusion Range is determined by degree of function even degree has a restricted range. It may not be easy to find but it is known to exist - if leading coefficient > 0 (pos)the range has a minimum – If the leading coefficient < 0 (neg) it flips the polynomial and the range has a maximum Odd degree functions have a range of all real numbers – the function goes neg. infinity on the left to positive infinity on the right. Essentially an increasing function Leading coefficient < 0 (neg) flips the polynomial making it go from positive infinity on the left to neg infinity on the right – essentially a decreasing function. The above statements are sometimes referred to as “End behavior”. This behavior is determined by degree and by the leading coefficient

Examples Determine degree for the following polynomials – what does this tell you about the graph? f(x) = 2x5 – 7x4 + 3x2 – 9 g(x) = (x + 2)(x – 5)( x + 3)2 k(x) = (x + 2)3 – (x + 3)2 + 5x – 7 m(x) = (x – 2)(3 – x) (2x – 5)2 z(x) = -3x7 – 4x5 + 2x3 Determining max/min cannot be done with algebra except by trial and error (making a table) this is not a desirable method and Calculus offers a way to find max and min points

Using a graphing calculator to estimate local max/min Ex- 9.1x4 -71.3x3+170.09x2-108.2x+45 Ex: (x – 3)(x – 6)(x + 4) [-8,8,1] x[-300,200,50]

Y- intercepts f(0) = y determines the y – intercept Ex: f(x) = 2x5 – 7x4 + 3x2 – 9 g(x) = (x + 2)(x – 5)( x + 3)2 k(x) = (x + 2)3 – (x + 3)2 + 5x - 7

X- intercepts- what’s happening in the middle of the graph Given f(x) = y , Solving f(x) = 0 determines the x-intercepts The solutions to this equation are often referred to as the zeroes of the function and by association the x – intercepts are also sometimes called the zeroes of the function One method of solving a polynomial equation is by factoring solutions - x - intercepts factors Conclusion: given f(x) x = a is a solution of f(x) = 0 if and only if (x –a) is a factor of f(x) and if and only if (a,0) is a point on the graph.

examples Determine the degree of each polynomial and the x-intercepts of the graph and sketch the graph g(x) = (x – 2)(x + 4)(x – 9)(5x + 6) k(x) = .1x(x + 5)(2x – 9)

Example Find all solutions to x3 – 64 = 0 sketch the graph for f(x) = x3 – 64 Find all solutions to x4 – 10x2 – 24 = 0 sketch the graph for g(x)= x4 – 10x2 – 24 Find all solutions to x5 – 9x3 + 8x2 – 72 = 0 sketch the graph for k(x) = x5 – 9x3 + 8x2 – 72

Multiplicity and x-intercept behavior When a solution occurs more than once it is said to “have multiplicity” (x – 5)2 yields a solution of 5 with multiplicity of 2 – the x-intercept (5,0) is said to have multiplicity of 2 (x - 5)7 solution of 5 with multiplicity of 7 The behavior of these intercepts corresponds to what we already know about power graphs

Multiplicity of factors j(x) = (-1/512)(x + 8)3(x – 7)2(x + 2) Odd multiplicity – Even multiplicity

Ex: Mult/intercepts Sketch a graph of the following polynomials f(x) = (x + 2)(x – 3)(x – 7) g(x) = (x + 2)2(x – 3)3 (x – 7) k(x) = -2(x + 2)(x – 3)(x – 7)2

Using long division to factor The factors of a polynomial are not always easy to find – Pertinent fact about factors : If a number is a factor of another number then it divides with no remainder

Long division/synthetic division You can always do long division You can only do synthetic division when you are dividing by a binomial of degree 1 with a coefficient of 1

Synthetic division It’s a lot of writing to write all the variables over and over again. Synthetic division is a “shorthand” for long division – (2x7 + 6x6 - 9x4 - 2x3 + 3x + 6)/ (x + 2) becomes: (notice the sign change and that every term is represented) -2| 2 6 0 -9 -2 0 3 6 Answer:

Example (x3 - 4x2 – 7x + 10)÷ (x + 2) (6x4 +10x3 – 19x2 – 25x +10)/(2x2 – 5)

Examples (x3 – 5x2 + 7x -6)÷ (x – 4) (x4 – 2x2 + 5x + 5)/ (x + 2)

Division/factoring/zeroes(solutions) Importance : if the remainder is zero then (x – a) is a FACTOR of the polynomial ie : we can use division to find factors that are not binomial factors Given x + 2 is a factor of the function p(x) = x3 - 4x2 – 7x + 10 then (x3 - 4x2 – 7x + 10)÷ (x + 2) = x2 - 6x +5 And x2 - 6x +5 is also a factor - therefore the factors of the polynomial are (x +2)(x -5)(x -1) If there are enough binomial factors we can determine ALL the solutions (rational, irrational, and imaginary) for the given polynomial

Example Given g(x) = X5 + 2x4 – 12x3 – 24x2 + 27x + 54 is x = -2 a solution? is x = 1 a solution? is (x + 3 ) a factor of g(x)? is (x +1) a factor of g(x)?

Rational factor theorem- what factors are possible 1) like the ac method of factoring a quadratic - any rational solution is determined by the factors of the leading coefficient and constant terms axn +………….+ c has rational solutions of 𝑐 1 𝑎 1 where c1 and a1 are factors of c and a 2) any rational factors yield a solution to the equation f(x) = 0 Thus (x – a) is a factor if f(a) = 0 3) any factor will divide nicely into the polynomial – thus yielding other factors

Example: x – intercepts vs solutions Ex: Given f(x) = 2x3 – x2 – 7x + 6 First - possible solutions 6 : 1,2,3,6 2: 1,2 there are at most 16 possible solutions Second - determine at least one actual solution f(1) f(-1) f(2) f(-2) …… Third – divide the polynomial by (x - ) and factor the resulting polynomial solutions are : x- intercepts are:

Examples Find all solutions for 3x3 - 5x2 - 34x – 24 = 0 Find all x- intercepts for g(x)= 3x3 - 5x2 - 34x – 24 sketch the graph Find all solutions for x4 + 2x3 – 11x2 + 8x = 60 Find all x- intercepts for k(x) = x4 + 2x3 – 11x2 + 8x-60

Summary domain/ range - (end behavior) y – intercepts x-intercepts - factors - multiplicity finding possible factors by ac rule testing to see if a binomial is a factor finding factors by division Use the above information to sketch a graph

Assignment P 7 -23(1-10odd)(15 -29 odd)(37-59 odd- use tracing on the word problems)(64 )(65 – 70 odd) worksheet

Rational functions Chapter 7 – section 2

Rational functions A rational function is the quotient of 2 polynomial functions 𝑓 𝑥 = 𝑝(𝑥) 𝑞(𝑥) where p(x) and q(x) are polynomial functions Ex:

Domain Domain of rational function are generally restricted - Denominator cannot equal zero This has already been discussed Review: find domain for 𝑓 𝑥 = 𝑥 2 −4 𝑥 2 −7𝑥−18 𝑔 𝑥 = 9 𝑥 2 − 6𝑥 + 4 j x = 𝑥 −5 𝑥 2 +9 Range may or may not be restricted – this is not easily determined and will not be discussed here

Key elements vertical asymptotes and “holes” A restriction in the domain means that there are NO points above those x-values - since these are single points (not intervals) the graph must “skip” that point. This either creates a “hole” or a “wall” (called an asymptote) Horizontal asymptotes: these occur when the quotient approaches a constant value for large values of x - they frequently cause a skip in the range but not always Oblique asymptotes

Vertical asymptotes –vs- holes If a factor in the denominator also occurs in the numerator the restriction at that value creates a hole Otherwise a restriction causes an asymptote Examples: 𝑓 𝑥 = 𝑥 2 −4 𝑥 2 −7𝑥−18 𝑔 𝑥 = 3𝑥 −7 𝑥 2 −2𝑥 −24

Horizontal asymptotes – end behavior Recall what happens when x is HUGE either negative or positive The behavior was dependent on the degree of the polynomial Now we are dividing – there are 3 possibilities

Conclusions Given 𝑓 𝑥 = 𝑝(𝑥) 𝑞(𝑥) where the degree of p(x) = a and the degree of q(x) = b If a < b the horizontal asymptote is y = 0 If a> b there is no horizontal asymptote (there is a possibility of a slant asymptote) If a = b the horizontal asymptote is the fraction created by the leading coefficients of numerator and denominator

examples 𝑓 𝑥 = 𝑥+5 3𝑥 2 −8𝑥 𝑘 𝑥 = 5 𝑥 2 −9 𝑥+2 𝑔 𝑥 = 7𝑥 −15 2𝑥 −6

Oblique asymptotes/ slant asymptotes When the numerator is the higher degree 𝑝(𝑥) 𝑞(𝑥) =𝑓 𝑥 + 𝑟(𝑥) 𝑞(𝑥) where f(x) is a polynomial function and r(x) is the remainder from dividing. f(x) defines the end behavior of the original function and if f(x) is linear is called a slant asymptote

examples 𝑘 𝑥 = 2 𝑥 4 −5 𝑥 2 2 𝑥 3 +5𝑥 6 𝑥 5 + 2 𝑥 4 −2𝑥 −15 3 𝑥 3 +4 𝑥 2

Rational graphs From previous chapters you should know what the graph of f x = 1 𝑥 looks like and any transformations on that graph Beyond that you should recognize a rational graph as a rational graph as opposed to a polynomial graph when you see one

Assignment P616(1-16) (28 – 32)