1. TBF 121 - General Mathematics - I Lecture – 4 : Polynomials and Rational Functions Prof. Dr. Halil İbrahim Karakaş Başkent University

2. Polynomial Functions. Let a 0, a 1,..., an an be real numbers. The function f defined by the equation is called a polynomial function or shortly a polynomial. The numbers a 0, a 1,..., an an are called the coefficients of the polynomial. a 0 : constant term, ai ai : i-th coefficient Zero polynomial : the polynomial with all coefficients equal to zero. f(x) f(x) = 0.0. The domain of every polynomial function is ℝ.ℝ. if a n ≠ 0, then is called the leading coefficient of f and n is called the degree of f. The degree of f is denoted by deg(f). Given a polynomial The degree of f the zero polynomial is defined to be ..

3. Special Polynomials. Constant function: f(x) = b, b b ℝ If c c ℝ and f(c) = 0, c is called a root of f. If f and g are polynomials and c is a real number such that f(x)=(x–c)g(x), then c is a root of f.f. Let f be a polynomial of degre n ≥ 1. If c  ℝ is a root of f, then there is a polynomial g of degree n–1 such that f(x)=(x–c)g(x). It follows that Linear function: f(x) = mx+b; m, b b ℝ, m ≠ 0. Quadratic Function: f(x) = ax 2 +bx+c; a, a, b, b, c  ℝ, a ≠ 0. Actually we have A nonzero polynomial of degre has at most n roots. Note that roots of a polynomial give its x-intercepts.

4. Long Division. x–1 3x23x2 3x 3 – 3x 2 5x 2 –7x–7x +5x 5x 2 – 5x –2x+2 -2 -2x+2 0

5. Let c  ℝ and k  ℕ.ℕ. f is continuous on (– ,  ).

6. For the polynomial f (x) = 3x5 3x5 +2x 2 –7x+2

7. Constant Functions. f(x) f(x) = b, b b ℝ.ℝ. x y (0,0) (–1,b) (0,b) (2,b) f (x) = b

8. Linear Functions. As a polynomial function the domain of a linear function is ℝ.ℝ. f(x) f(x) = mx + b ; m, b  ℝ, m ≠ 0. What is the image of a linear function? The identity function f(x) f(x) = x is a linear function: m=1 and b=0. x y (0,0) y = x x y (0,0) y = mx x y (0,0) y = mx+ b (0,b) The graph of any linear function is a line.

9. The graph of the linear function f(x) f(x) = 2x + 4 : x-intercept : f(x) f(x) = 0  x = - 2  (-2, 0). y-intercept : (0, f(0)) = (0, 4) x y (0,0) (-2, 0) (0, 4) f(x) = 2x + 4

10. The graph of f(x)=–2x : Note that f(x) = 0  x= x= 0. x-intercept and y-intercept coincide: (0,0). x = 1.  f(1)=–2  (1,–2) is on the graph.. x y (0,0) (1,-2) f(x) = -2x

11. Example. The total cost of x units of an item is a linear function of x. At one time 100 units of that item cost 200 TL, at another time 150 units cost 275 TL. Find the equation that defines the cost function C.C. C(x) = mx + b C(100) = 100m + b = 200 and C(150) = 150m + b = 275 From the first equation we get b = 200–100m and plug it in the second equation 150m + (200–100m) = 275 ⇨ 50m = 75 ⇨ m = 1.5. So b = 50 and C(x) = (1.5)x + 50.

12. Example. A car dealer plans to launch a new model car. He decides to start the launch if the price of a car is over 30000 TL. Besides, for each 1000 TL increase in the price he plans to put 5 more cars for sale. Price – suply function is known to be linear. a) Find the equation defining the price – suply function. b) How many cars will be on sale when the price is 35000 TL ? a)p(x) = mx + b p(x) = 200x + 30000. b = p(0) = 30000 p(5) = 5m + b = + 30000=31000  m =200 b) 35000 = 200x + 30000  x = 25.  p(x) = mx + 30000

13. Lines in the Plane. x y (0,0) (x,b)(x,b) (0,b) Horizontal Line : y = b constant function x y (0,0) (a,0) (a,y)(a,y) Vertical Line : x= a Not a function!

14. x y (0,0) (0,b) Using similar triangles Inclined Line d slope of dSlope - Intercept FormPoint - Slope Form

15. Slope - Intercept FormPoint - Slope Form Line with slope m = 3 and y – intercept b = 4:4: Line with slope m = 3 and passing through (1,2): If two points (x1 (x1, y1 y1 ), ( x2 x2, y2 y2 ) of a line are known, the slope of the line is computed as And the equation of the line can be written in sıope-intercept form. Line passing through the points (1,2) and (3,5):

16. Linear Equations. Ax + By = C İs called a linear equation. A and B are called coefficients, C is called the right hand side constant of the linear equation. The symbols x and y are called the variables. Graph of Ax + By = C : Linear function, inclined line Constant function, horizontal line Not a function, vertical line Let A, B, C be real numbers such that A  0 or B  0. The equation

17. Quadratic Functions. f(x) = ax 2 + bx + c is called a quadratic function. Completing to the square If we let we obtain Let a, b, c be real numbers such that a a 0. The function f defined by the equation

18. Summarizing what we have obtained, The graph of the quadratic function is a parabola. Horizontal shift Strech, shrink, reflection Vertical shift

19. Let us skech the graph of f(x) f(x) = ax 2 + bx + c when a > 1, h > 0, k > 0 : x y (0,0) x y x y (h,0) (h,k) minimum value of f is f(h) = k. Horizontal shift to the right Strech Vertical shift upwards

20. Let us skech the graph of f(x) f(x) = ax 2 + bx + c when a <, h > 0, k > 0 : x y (0,0) x y x y (h,0) (h,k) maksimum value of f is f(h) = k Horizontal shift to the right Strech and reflection about x-axis Vertical shift upwards

21. f(x) = ax 2 + bx + c = a(x-h)2 a(x-h)2 + k, f(h) = k The point (h,k) is called the vertex of the parabola which is the graph of the quadratic function f(x) = ax 2 + bx + c = a(x-h)2 a(x-h)2 + k.k. Thus the shape of the quadratic function f(x) = ax 2 + bx + c is determined by the sign of a,a, the vertex, the x-intercepts and the y-intercept. If a > 0, f(h) = k is the minimum value of f(x); f(x) has no maximum value in this case. If a < 0, f(h) = k is the maximum value of f(x) ); f(x) has no minimum value in this case. When a > 0 is the lowest point of the parabola and the parabola opens upwards. When a < 0 is the highest point of the parabola and the parabola opens downwards.

22. Example. x y (0,0) (6,3) x-intercept : none y-intercept : f (0) = 21, (0, 21) (0,21) Vertex : (6, 3) Parabola upwards ( a > 0)

23. Example. x y (0,0) (4,8) x-intercepts : f (x) = 0  -2x 2 + 16x –24 = 0  x = 2, 6  (2, 0), (6, 0) y-intercept : f (0) =- 24, (0, -24) (0,-24) Vertex : (4, 8) Parabola downwards( a < 0) (2, 0) (6, 0)

24. Applications of Linear and Quadratic Functions. Solution. Let the number of subscribers be 1200 + x, x  0. Then the profit of the company is This quadratic function reaches its maximum value for Example. A telephone company has a profit of 20 TL per subscriber if the number of subscribers is less than or equal to 1200. After 1200, each new subscriber causes a decrease of 0.01 TL in the profit. What should be the number of subscribers for the profit of the company be maximum? What is the maximum profit? Thus the profit is maximum for 1200 + 400 = 1600 subscribers. Maximum profit is P(400)=25600 TL.

25. Solution. Example. A company that manufactures teapots can sell x = 4000–20p teapots in one year for a price of p TL per teapot. If the company aims to have a revenue of 150000 TL at the end of the year, how many teapots should be sold and what should be the price of a teapot? The revenue when x teapots are sold: R(x)=xp=(4000–20p)p=4000p–20p 2 Thus for a revenue of 150000 TL,  (p–150)(p–50) = 0 150000 = 4000p–20p 2  p 2 –200p +7500 = 0 The revenue will be 150000 TL if the price is 50 TL per teapot and x = (4000–20. 50)= 3000 teapots are sold or the price is 150 TL per teapot and x = (4000–20. 150) = 1000 teapots are sold.  p = 50 or p = 150.

26. Example. The revenue and cost functions for a company are given as thousand TL, where x denotes the number of items produced and sold. Sketch the graphs of R and C on the same coordinate plane and answer the following questions: a) Find the level of production for which the cost and the revenue are equal (break even). b) Find the values of x for which the company will have a profit and the values of x for which the company will have a loss. Solution. The revenue function is a quadratic function For break even:

27. 10 200 400 y x (0,0) 1 15 6 12 PROFİT loss revenue = cost (break even) Break even for 6 or 12 items. Profit in the interval (6,12), that is, the company will have a profit if the number of items produced and sold is more than 6 or less than 12; and a loss in the intervals (1,6) and (12,15).

28. Look at the profit function: Maximum profit : 36000 TL. (thousand TL).

29. Rational Functions. where p(x) ande d(x) are polynomials. The polynomials p(x) p(x) and d(x) d(x) are called the numerator and the denominator of f, respectively. We assume that in the above expression (*), the numerator and the denomianator have no nonconstant common factor.. The domain of the rational function defined in (*) is the set { x : d(x)  0 } In sketching the graph of a rational function f, it is important and helpful to know how f (x) changes as x   or x  - . It is also important and helpful to know how f (x) changes as x  a - or x  a+ a+ for each a  ℝ with d(a) d(a) = 0.0. A rational function is a function f defined by an equation of the form (*) x = a is a vertical asymptote if d(a) d(a) = 0.0. y = b is an horizontal asymptote if f (x)  b as x   or x  - 

30. , let c  ℝ, d(c)  0. Then f is continuous at x = c. Thus f is continuous throughout its domain Example. Domain : { x : x  0 } = ℝ \{0} = (- , 0)  (0, )) Horizontal asymptote : y = 0.0. Vertical asymptote : x = 0.0.

31. Example. Domain : { x : x+1  0 } = ℝ \{-1} = (- , -1)  (-1, )) Similarly, y = 1 is horizontal asymptote of g. On the other hand, x = is vertical asymptote of g.

32. The graph of can be obtained from the graph of the hyperbolic function by elementary transformations: –1–1 1 y x (0,0) x y (0,–2) (2,0)

33. Limits of Rational Functions at infinity.

34. Examples.

35. Example. As x  –  or x  , we have y  0. So y = 0 is horizontal asymptote. As x  –2 – or x  2– 2–, we have y  – – ; as x  –2 + or x  2 +, we have y   So x =–2 and x = 2 are vertical asymptotes.. The graph looks like the following: x y (0,0) -2 2

36. Application (On-the-job training). An electronics company has established that, on the average, a new employee can assemble N(t) components per day after t days of on-the-job training where N(t) is given by Graph y = N(t) and interpret it. Vertical asymptote: t + 4 = 0  t = -4 -4. As t  - 4 - iken s(t)   t  - 4+ 4+, N(t)  -  We are interested in that part of the graph where t  0. Horizontal asymptote: As t  -  or t  , N(t)  40  y = The graph is on the next slide.

37. t y (0,0) -4 50 As the number of training days increases, the number of components assembled also increases; but after some days (196 days) this number stabilizes.

38. (1,1) (-1,1) Piecewise – defined functions. There are functions which are defined by different equations or rules over different intervals. The absolute value function is an example: This expression shows that the absolute value function is defined by the equation |x| = x over the interval [0,  ) while it is defined by the equation |x|= - x over the interval (- ,0). A function which is defined by different equations or rules over different intervals is called a piecewise – defined function. The graph of a picewise – defined function is obtained by sketching the graph each equation in the definition over the interval that it belongs to. Example. The graph of the piecewise – defined function (2,0) (-2,0) y x

39. Example. A textile company decides the wholesale price of a meter of a kind of cloth as follows: A customer that buys 10 meters or less pays 10 TL per meter; if the customer buys more, after paying 10 TL per meter for the first 10 meters, he pays 8 Tl per meter for the next 40 meters and if he buys more, he pays 6 TL per meter for the rest. Write a piecewise definition of the amount f(x) that a customer will pay if he/she buys x meters of cloth. If x > 50, for the first 10 meters, 10. 10=100 TL ; for the next 40 meters, 8. 40=320 TL; and for the remaining (x – 50) meters, 6(x – 50) TL will be paid. So, if x > 50, then we have f (x) = 100 + 320 + 6(x – 50) = 6x + 120 TL. If 0 < x ≤ 10, then f (x) = 10x TL. If 10 < x ≤ 50, 10. 10 = 100 TL will be paid for the first 10 meters, and for the remaining (x – 10) meters, 8(x – 10) TL will be paid. So f (x) = 100 +8(x – 10) = 8x + 20 TL. Hence