End Behavior of Functions

1. End Behavior of Functions The end behavior of a graph describes the far left and the far right portions of the graph. End behavior: A description of what happens to the values f(x) of a function f as x  ∞ and as x  -∞ Using the leading coefficient and the degree of the polynomial, we can determine the end behaviors of the graph. This is often called the Leading Coefficient Test.

2. Smooth, Continuous Graphs Two important features of the graphs of polynomial functions are that they are smooth and continuous. By smooth, we mean that the graph contains only rounded curves with no y y sharp corners. By continuous, we mean that the graph has no breaks and can be drawn without lifting your pencil from the rectangular coordinate system. These ideas are illustrated in the figure. Smooth rounded corner Smooth rounded corner x x Smooth rounded corner Smooth rounded corner

3. As xincreases or decreases without bound, the graph of the polynomial function f (x)=anxn+ an-1xn-1+ an-2xn-2 +…+a1x + a0 (an¹ 0) eventually rises or falls. In particular, 1. For n odd: an> 0an< 0 If the leading coefficient is positive, the graph falls to the left and rises to the right. If the leading coefficient is negative, the graph rises to the left and falls to the right. Rises right Rises left Falls right Falls left The Leading Coefficient Test

4. As xincreases or decreases without bound, the graph of the polynomial function f (x)=anxn+ an-1xn-1+ an-2xn-2 +…+a1x + a0 (an¹ 0) eventually rises or falls. In particular, 1. For n even: an> 0an< 0 If the leading coefficient is positive, the graph rises to the left and to the right. If the leading coefficient is negative, the graph falls to the left and to the right. Rises right Rises left Falls left Falls right The Leading Coefficient Test

5. End Behavior • Four Possibilities • Up on both ends • Down on both ends • Up on the right & Down on the left • Up on the left & Down on the right

6. End Behavior Notation: • Up on both ends… • Down on both ends… • Up on the right & Down on the left… • Up on the left & Down on the right…

7. End Behavior degree • If the __________ is even and the leading coefficient is _________, then positive degree B. If the __________ is even and the leading coefficient is _________, then negative

8. End Behavior degree C. If the __________ is odd and the leading coefficient is _________, then positive degree D. If the __________ is odd and the leading coefficient is _________, then negative

9. End Behavior of Functions First determine whether the degree of the polynomial is even or odd. degree = 2 so it is even Next determine whether the leading coefficient is positive or negative. Leading coefficient = 2 so it is positive

10. END BEHAVIOR Degree: Even Leading Coefficient: + End Behavior: UpUp

11. END BEHAVIOR Degree: Even Leading Coefficient: End Behavior: DownDown

12. END BEHAVIOR Degree: Odd Leading Coefficient: + End Behavior: Down Up

13. END BEHAVIOR Degree: Odd Leading Coefficient: End Behavior: UpDown

14. END BEHAVIOR Give the End Behavior: Up Down Up Up Down Up Down Down

15. keys to sketching a graph of a polynomial are: f(x) = 6x4 + x3 – 12 Leading coefficient Degree (highest exponent) (in front when in standard form!) Is it an even or odd number? (not in the sense of symmetry though. To find even/odd symmetry you have to plug in –x) Is it positive or negative?

16. Multiplicity of Real Zeros (how many times will it cross the x-axis): f(x) = 6x4 + x3 – 12 Degree (highest exponent) This will determine how many times it will cross the x-axis NOTE: sometimes it crosses the same number more then once

17. Rises right y x Falls left Text Example Use the Leading Coefficient Test to determine the end behavior of the graph of Graph the quadratic functionf(x)= x3+3x2-x- 3. SolutionBecause the degree is odd (n=3) and the leading coefficient, 1, is positive, the graph falls to the left and rises to the right, as shown in the figure.

18. Multiplicity and x-Intercepts • Ifris a zero of even multiplicity, then the graph touches the x-axis and turns around at r. • If r is a zero of odd multiplicity, then the graph crosses the x-axis at r. • Regardless of whether a zero is even or odd, graphs tend to flatten out at zeros with multiplicity greater than one.

19. Example • Find the x-intercepts and multiplicity of f(x) = 2(x+2)2(x-3) Solution: • x=-2 is a zero of multiplicity 2 or even • x=3 is a zero of multiplicity 1 or odd

20. Turning Points of Polynomial Functions The graph of f(x) = x5-6x3 + 8x + 1 is shown in the figure below. The graph has four smooth turning points. At each turning point, the graph changes direction from increasing to decreasing or vice versa. In calculus, these points are called local maxima or local minima. The given equation has 5 as its greatest exponent and is therefore a polynomial function of degree 5. Notice that the graph has four turning points. In general, if f isa polynomial ofdegree n, then the graph offhas at most n- 1 turning points. Turning points: from increasing to decreasing. y x Turning points: from decreasing to increasing.

21. Graphing a Polynomial Function f (x)=anxn+ an-1xn-1+ an-2xn-2 +  +a1x + a0 (an 0) 1. Use the Leading Coefficient Test to determine the graph's end behavior. 2. Find x-intercepts by settingf(x) = 0 and solving the resulting polynomial equation. If there is an x-intercept at r as a result of (x-r)kin the complete factorization of f(x), then: a. If k is even, the graph touches the x-axis at r and turns around. b. If k is odd, the graph crosses the x-axis at r. c. If k > 1, the graph flattens out at (r, 0). 3.Find the y-intercept by setting x equal to 0 and computing f(0).

22. f (x)=anxn+ an-1xn-1+ an-2xn-2 +  +a1x + a0 (an 0) Graphing a Polynomial Function 4. Use symmetry, if applicable, to help draw the graph: a.y-axis symmetry: f(-x) =f(x) b.Origin symmetry: f(-x) =-f(x). 5. Use the fact that the maximum number of turning points of the graph is n - 1 to check whether it is drawn correctly.

23. Solution Step 1 Determine end behavior. Because the degree is even (n = 4) and the leading coefficient, 1, is positive, the graph rises to the left and the right: y Rises left Rises right x Text Example Graph: f(x)=x4-2x2+ 1.

24. Text Example cont. Graph: f(x)=x4-2x2+ 1. Solution Step 2 Find the x-intercepts (zeros of the function) by settingf(x)=0. x4-2x2+ 1 = 0 (x2- 1)(x2- 1) = 0 Factor. (x+ 1)(x -1)(x+ 1)(x -1) = 0 Factor completely. (x+ 1)2(x- 1)2= 0 Express the factoring in more compact notation. (x+ 1)2= 0 or (x- 1)2= 0 Set each factor equal to zero. x=-1 x= 1Solve for x.

25. y Rises right Rises left x 1 1 Text Example cont. Graph: f(x)=x4-2x2+ 1. Solution Step 2We see that -1 and 1 are both repeated zeros with multiplicity 2. Because of the even multiplicity, the graph touches the x-axis at -1 and 1 and turns around. Furthermore, the graph tends to flatten out at these zeros with multiplicity greater than one:

26. y Rises right Rises left 1 x 1 1 Text Example cont. Graph: f(x)=x4-2x2+ 1. Solution Step 3 Find the y-intercept. Replace xwith 0 in f(x)=-x+ 4x - 1. f(0) =04- 2 • 02+ 1=1 There is a y-intercept at 1, so the graph passes through (0, 1).

27. Replace xwith -x. Graph: f(x)=x4-2x2+ 1. Solution Step 4 Use possible symmetry to help draw the graph. Our partial graph suggests y-axis symmetry. Let's verify this by finding f(-x). f(x) =x4-2x2+ 1 f(-x)= (-x)4-2(-x)2+ 1 =x4-2x2+ 1 Because f(-x)= f(x), the graph of f is symmetric with respect to the y-axis. The following figure shows the graph.

28. y x Graph: f(x)=x4-2x2+ 1. Solution Step 5 Use the fact that the maximum number of turning points of the graph is n- 1 to check whether it is drawn correctly. Because n= 4, the maximum number of turning points is 4 - 1, or 3. Because our graph has three turning points, we have not violated the maximum number possible.

29. y x Text Example cont. Graph: f(x)=x4-2x2+ 1. Solution