Rational Functions A rational function f(x) is a function which is the ratio of two Elementary Functions polynomials, that is, Part 2, Polynomials f(x) = n(x) Lecture 2.6a, Rational Functions d(x) where n(x) and d(x) are polynomials. Dr. Ken W. Smith 3x2−x−4 For example, f(x) = x2−2x−8 is a rational function. Sam Houston State University In this case, both the numerator and denominator are quadratic 2013 polynomials. Smith (SHSU) Elementary Functions 2013 1 / 42 Smith (SHSU) Elementary Functions 2013 2 / 42 Algebra with mixed fractions Algebra with mixed fractions Consider the function g(x) which appeared in an earlier lecture: 1 2x − 3 (2x + 1) + (2x − 3)(x + 2) + (x − 5)(2x + 1)(x + 2) g(x) := + + x − 5: g(x) = : x + 2 2x + 1 (2x + 1)(x + 2) This function, g, is a rational function. We can put g into a fraction form, as the ratio of two polynomials, by finding a common denominator. The numerator is a polynomial of degree 3 (it can be expanded out to 2x3 − 3x2 − 20x − 15) and the denominator is a polynomial of degree 2. The least common multiple of the denominators x + 2 and 2x + 1 is simply their product, (x + 2)(2x + 1): We may write g(x) as a fraction with this The algebra of mixed fractions, including the use of a common 2x+1 denominator if we multiply the first term by 1 = 2x+1 , multiply the second denominator, is an important tool when working with rational functions. x+2 (2x+1)(x+2) term by 1 = x+2 and multiply the third term by 1 = (2x+1)(x+2) : Then 1 (2x + 1) 2x − 3 (x + 2) (2x + 1)(x + 2) g(x) = ( ) + ( ) + (x − 5) : x + 2 (2x + 1) 2x + 1 (x + 2) (2x + 1)(x + 2) Combine the numerators (since there is a common denominator): (2x + 1) + (2x − 3)(x + 2) + (x − 5)(2x + 1)(x + 2) g(x) = : (2x + 1)(x + 2) Smith (SHSU) Elementary Functions 2013 3 / 42 Smith (SHSU) Elementary Functions 2013 4 / 42 Zeroes of rational functions Algebra with mixed fractions n(x) Given a rational function f(x) = d(x) we are interested in the y- and x- intercepts. The y-intercept occurs where x is zero and it is usually very easy to For example, suppose compute x2−6x+8 h1(x) = x2+x−12 : n(0) f(0) = d(0) : 2 8 2 The y-intercept is (− 3 ; 0) since h1(0) = −12 = − 3 : However, the x-intercepts occur where y = 0, that is, where The x-intercepts occur where x2 − 6x + 8 = 0: n(x) 2 0 = d(x) : Factoring x − 6x + 8 = (x − 4)(x − 2) tells us that x = 4 and x = 2 should be zeroes and so (4; 0) and (2; 0) are the x-intercepts. As a first step to solving this equation, we may multiply both sides by d(x) (We do need to check that they do not make the denominator zero { but and so concentrate on the zeroes of the numerator, solving the equation they do not.) 0 = n(x): At this point, we have reduced the problem to finding the zeroes of a polynomial, exercises from a previous lecture! Smith (SHSU) Elementary Functions 2013 5 / 42 Smith (SHSU) Elementary Functions 2013 6 / 42 Poles and holes Poles and holes The domain of a rational function is all the real numbers except those which make the denominator equal to zero. Since rational functions have a denominator which is a polynomial, we There are two types of zeroes in the denominator. must worry about the domain of the rational function. In particular, any One common type is a zero of the denominator which is not a zero of the real number which makes the denominator zero cannot be in the domain. numerator. In that case, the real number which makes the denominator zero is a \pole" and creates, in the graph, a vertical asymptote. The domain of a rational function is all the real numbers except those x2−6x+8 which make the denominator equal to zero. For example For example, using h1(x) = x2+x−12 from before, we see that x2 + x − 12 = (x + 4)(x − 3) has zeroes at x = −4 and x = 3: x2−6x+8 (x−4)(x−2) h1(x) = x2+x−12 = (x+4)(x−3) Since neither x = −4 and x = 3 are zeroes of the numerator, these values has domain (−∞; −4) [ (−4; 3) [ (3; 1) since only x = −4 and x = 3 give poles of the function h1(x) and in the graph we will see vertical lines make the denominator zero. x = −4 and x = 3 that are \approached" by the graph. The lines are called asymptotes, in this case we have vertical asymptotes with equations x = −4 and x = 3: Smith (SHSU) Elementary Functions 2013 7 / 42 Smith (SHSU) Elementary Functions 2013 8 / 42 Poles and holes Algebra with mixed fractions We continue to look at 2 h (x) = x −6x+8 = (x−4)(x−2) 1 x2+x−12 (x−3)(x+4) If we change our function just slightly, so that it is The figure below graphs our function in blue and shows the asymptotes. x2−6x+8 (x−4)(x−2) (The graph is in blue; the asymptotes, which are not part of the graph, are h2(x) = x2−x−12 = (x−4)(x+3) in red.) something very different occurs. The rational function h2(x) here is still undefined at x = 4: If one 0 attempts to evaluate h2(4) one gets the fraction 0 which is undefined. But, as long as x is not equal to 4, we can cancel the term x − 4 occurring both in the numerator and denominator and write x−2 h2(x) = x+3 ; x 6= 4: Smith (SHSU) Elementary Functions 2013 9 / 42 Smith (SHSU) Elementary Functions 2013 10 / 42 Poles and holes Poles and holes In this case there is a pole at x = −3, represented in the graph by a Here is the graph of vertical asymptote (in red) and there is a hole (\removable singularity") at x − 2 y = : x = 4 where, (for just that point) the function is undefined. x + 3 Smith (SHSU) Elementary Functions 2013 11 / 42 Smith (SHSU) Elementary Functions 2013 12 / 42 Here is the graph of (x − 4)(x − 2) (x − 4)(x + 3) The sign diagram of a rational function The sign diagram of a rational function Just as we did with polynomials, we can create a sign diagram for a When we looked at graphs of polynomials, we viewed the zeroes of the rational function. polynomial as dividers or fences, separating regions of the x-axis from one another. In this case, we need to use both the zeroes of the rational function and the vertical asymptotes as our dividers, our \fences" between the sign Within a particular region, between the zeroes, the polynomial has a fixed changes. sign, (+) or (−), since changing sign requires crossing the x-axis. To create a sign diagram of rational function, list all the x-values which We used this idea to create the sign diagram of a polynomial, a useful give a zero or a vertical asymptote. Put them in order. Then between tool to guide us in the drawing of the graph of the polynomial. these x-values, test the function to see if it is positive or negative and indicate that by a plus sign or minus sign. Smith (SHSU) Elementary Functions 2013 13 / 42 Smith (SHSU) Elementary Functions 2013 14 / 42 The sign diagram of a rational function Rational Functions For example, consider the function x2−6x+8 (x−4)(x−2) h2(x) = x2−x−12 = (x−4)(x+3) from before. It has a zero at x = 2 and a vertical asymptote x = −3: The sign diagram represents the values of h2(x) in the regions divided by In the next presentation we look at the end behavior of rational functions. x = −3 and x = 2: (For the purpose of a sign diagram, the hole at x = 4 is irrelevant since it does not effect the sign of the rational function.) To (END) the left of x = −3, h2(x) is positive. Between x = −3 and x = 1, h2(x) is negative. Finally, to the right of x = 1, h2(x) is positive. So the sign diagram of h2(x) is (+) j (−) j (+) −3 1 Smith (SHSU) Elementary Functions 2013 15 / 42 Smith (SHSU) Elementary Functions 2013 16 / 42 End-behavior of Rational Functions Just as we did with polynomials, we ask questions about the \end behavior" of rational functions: what happens for x-values far away from Elementary Functions 0, towards the \ends" of our graph? Part 2, Polynomials In many cases this leads to questions about horizontal asymptotes and Lecture 2.6b, End Behavior of Rational Functions oblique asymptotes (sometimes called \slant asymptotes"). Before we go very far into discussing end-behavior of rational functions, we Dr. Ken W. Smith need to agree on a basic fact. Sam Houston State University As long as a polynomial p(x) has degree at least one (and so was not just 2013 a constant) then as x grows large p(x) also grows large in absolute value. 1 So, as x goes to infinity, p(x) goes to zero. Smith (SHSU) Elementary Functions 2013 17 / 42 Smith (SHSU) Elementary Functions 2013 18 / 42 End-behavior of Rational Functions End-behavior of Rational Functions We explicitly list this as a lemma, a mathematical fact we will often use.