Honors Algebra 2

1. Honors Algebra 2 End Behavior and Graphing Polynomials

2. Quiz y=3x + 4 • Function? Yes or No • How did you decide your answer for #1? • State Domain • State Range

3. Warm Up Identify each function as linear or quadratic: 1. 2.

4. 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 Smooth, Continuous Graphs

5. End Behavior Activity

6. 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

7. 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

8. y y x x End Behavior of a Polynomial Function’s Graph

9. y y x x End Behavior of a Polynomial Function’s Graph

10. Rises right y x Falls left 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.

11. As x –∞, P(x) +∞, and as x+∞, P(x) –∞. Example 2A: Using Graphs to Analyze Polynomial Functions Identify whether the function graphed has an odd or even degree and a positive or negative leading coefficient. P(x) is of odd degree with a negative leading coefficient.

12. As x –∞, P(x) +∞, and as x+∞, P(x) +∞. Example 2B: Using Graphs to Analyze Polynomial Functions Identify whether the function graphed has an odd or even degree and a positive or negative leading coefficient. P(x) is of even degree with a positive leading coefficient.

13. As x –∞, P(x) +∞, and as x+∞, P(x) –∞. Check It Out! Example 2a Identify whether the function graphed has an odd or even degree and a positive or negative leading coefficient. P(x) is of odd degree with a negative leading coefficient.

14. Now that you have studied factoring, solving polynomial equations, and end behavior, you can graph a polynomial function.

15. Example 3: Graphing Polynomial Functions Graph the function. f(x) = x3 + 4x2 + x – 6. Step 1Identify the possible rational roots by using the Rational Root Theorem. ±1, ±2, ±3, ±6 p = –6, and q = 1. Step 2 Test all possible rational zeros until a zero is identified. Test x = –1. Test x = 1. –1 1 1 4 1 –6 1 4 1 –6 –1 –3 2 1 5 6 1 1 3 5 –2 –4 6 0 x = 1 is a zero, and f(x) = (x – 1)(x2 + 5x + 6).

16. 5 5 2 2 f(0) = –6, so the y-intercept is –6. Plot points between the zeros. Choose x = – , and x = –1 for simple calculations. f() = 0.875, and f(–1) = –4. Example 3 Continued Step 3Write the equation in factored form. Factor: f(x) = (x – 1)(x + 2)(x + 3) The zeros are 1, –2, and –3. Step 4Plot other points as guidelines.

17. x –∞, P(x) –∞, and as x+∞, P(x) +∞. Example 3 Continued Step 5Identify end behavior. The degree is odd and the leading coefficient is positive so as Step 6Sketch the graph of f(x) = x3 + 4x2 + x – 6 by using all of the information about f(x).

18. Example 3a Graph the function. f(x) = x3 – 2x2 – 5x + 6. Step 1Identify the possible rational roots by using the Rational Root Theorem. ±1, ±2, ±3, ±6 p = 6, and q = 1. Step 2 Test all possible rational zeros until a zero is identified. Test x = –1. Test x = 1. –1 1 1 –2 –5 6 1 –2 –5 6 –1 3 2 1 –1 –6 1 1 –3 –1 –2 8 –6 0 x = 1 is a zero, and f(x) = (x – 1)(x2 – x – 6).

19. Example 3a Continued Step 3Write the equation in factored form. Factor: f(x) = (x – 1)(x + 2)(x – 3) The zeros are 1, –2, and 3. Step 4Plot other points as guidelines. f(0) = 6, so the y-intercept is 6. Plot points between the zeros. Choose x = –1, and x = 2 for simple calculations. f(–1) = 8, and f(2) = –4.

20. x –∞, P(x) –∞, and as x+∞, P(x) +∞. Example 3a Continued Step 5Identify end behavior. The degree is odd and the leading coefficient is positive so as Step 6Sketch the graph of f(x) = x3 – 2x2 – 5x + 6 by using all of the information about f(x).

21. You Try Graph the function. f(x) = –2x2 – x + 6. Step 1Identify the possible rational roots by using the Rational Root Theorem. ±1, ±2, ±3, ±6 p = 6, and q = –2. Step 2 Test all possible rational zeros until a zero is identified. Test x = –2. –2 –2 –1 6 4 –6 –2 3 0 x = –2 is a zero, and f(x) = (x + 2)(–2x + 3).

22. The zeros are –2, and . 3 2 You Try Continued Step 3The equation is in factored form. Factor: f(x) = (x + 2)(–2x + 3). Step 4Plot other points as guidelines. f(0) = 6, so the y-intercept is 6. Plot points between the zeros. Choose x = –1, and x = 1 for simple calculations. f(–1) = 5, and f(1) = 3.

23. x –∞, P(x) –∞, and as x+∞, P(x) –∞. You Try Continued Step 5Identify end behavior. The degree is even and the leading coefficient is negative so as Step 6Sketch the graph of f(x) = –2x2 – x + 6 by using all of the information about f(x).

24. Check Yourself