1. Evaluate and Graph Polynomial Functions Section 2.2 How do you identify and evaluate polynomial functions? What is synthetic substitution? How do you graph polynomial functions?

2. Polynomial Function f(x) = a n x n + a n−1 x n−1 + … + a 1 x + a 0 a n ≠ 0 Exponents are all whole numbers Coefficients are all real numbers a n is the leading coefficient a 0 is the constant term n is the degree f(x) = 7x 3 + 9x 2 + 3x + 1 f(x) = a 3 x 3 + a 3−1 x 3−1 + a 3−2 x 3−2 +1

3. Polynomial Functions Degree TypeStandard form 0 Constant f(x) = a 0 1 Linear f(x) = a 1 x + a 0 2 Quadratic f(x) = a 2 x 2 + a 1 x + a 0 3 Cubic f(x) = a 3 x 3 + a 2 x 2 + a 1 x + a 0 4 Quarticf(x) = a 3 x 3 + a 3 x 3 + a 2 x 2 + a 1 x + a 0

4. Identify polynomial functions 4 Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient. a. h (x) = x 4 – x 2 + 3 1 SOLUTION a. The function is a polynomial function that is already written in standard form. It has degree 4 (quartic) and a leading coefficient of 1.

5. To Be Or Not To Be Is the function a polynomial function? If so, write in standard form, state the degree, type and leading coefficient. f(x) = ½x 2 − 3x 4 −7 f(x) = x 3 +3 x f(x) = 6x 2 + 2x −1 + x f(x) = −0.5x + πx 2 −√2

6. Synthetic Substitution Use synthetic substitution to evaluate f(x) = 2x 4 −8x 2 + 5x −7 when x = 3

7. Direct Substitution vs Synthetic Substitution Direct substitution is replacing a variable with a number in a function to find the value of that function at that point. If x = 3 what is the value of y (or f(x)). Synthetic substitution is another way to evaluate a polynomial function that involves fewer operations than direct substitution.

8. Direct Substitution Example Use direct substitution to evaluate f (x) = 2x 4 – 5x 3 – 4x + 8 when x = 3. f (x) = 2x 4 – 5x 3 – 4x + 8 f (3) = 2(3) 4 – 5(3) 3 – 4(3) + 8 = 162 – 135 – 12 + 8 = 23 Write original function. Substitute 3 for x. Simplify

9. Try this one… 5. g (x) = x 3 – 5x 2 + 6x + 1; x = 4 SOLUTION g (x) = x 3 – 5x 2 + 6x + 1; x = 4 Write original function. g (x) = 4 3 – 5(4) 2 + 6(4) + 1 Substitute 4 for x. = 64 – 80 + 24 + 1 = 9 Evaluate powers and multiply. Simplify Be careful when you put this into your calculator, especially when you change between adding/subtracting and multiplying. You may find it is easier to put parts in and record the values, then add/subtract in the last step.

10. Synthetic Substitution Use synthetic substitution to evaluate f (x) from Example 2 when x = 3. f (x) = 2x4 – 5x3 – 4x + 8 SOLUTION STEP 1 Write the coefficients of f (x) in order of descending exponents. Write the value at which f (x) is being evaluated to the left.

11. STEP 2Bring down the leading coefficient. Multiply the leading coefficient by the x-value. Write the product under the second coefficient. Add. STEP 3Multiply the previous sum by the x-value. Write the product under the third coefficient. Add. Repeat for all of the remaining coefficients. The final sum is the value of f(x) at the given x-value.

12. Try this one… Use synthetic substitution to evaluate the polynomial function for the given value of x. 6. f (x) = 5x 3 + 3x 2 – x + 7; x = 2 STEP 1 Write the coefficients of f (x) in order of descending exponents. Write the value at which f (x) is being evaluated to the left.

13. STEP 2Bring down the leading coefficient. Multiply the leading coefficient by the x-value. Write the product under the second coefficient. Add. STEP 3Multiply the previous sum by the x-value. Write the product under the third coefficient. Add. Repeat for all of the remaining coefficients. The final sum is the value of f(x) at the given x-value. Synthetic substitution gives f(2) = 57 ANSWER

14. Synthetic substitution gives f(3) = 23, which matches the result in Example 2. ANSWER

15. Try this one… 7. g (x) = – 2x 4 – x 3 + 4x – 5; x = – 1 – 1 – 2 – 1 0 4 – 5 2 –1 1 –10 – 1 – 2 – 1 0 4 – 5 – 2 1 –1 5 – 10 Synthetic substitution gives f(– 1) = – 10 ANSWER

16. 2.2 Assignment Day 1 Page 99, 5-8, 10-12, 15-23 odd

17. End Behavior The end behavior of a polynomial graph is what happens to the graph as it continues past the physical graph. There are arrows on the ends of the graph that shows the direction in which the graph continues. The graph approaches positive infinity (+∞) or negative infinity (−∞). x→+∞ is read as “x approaches positive infinity.

18. The degree is odd and leading coefficient is negative. So, f (x) → + ∞ as x → – ∞ and f (x) → – ∞ as x → + ∞. The degree is even and leading coefficient is positive. So, f (x) → ∞ as x → – ∞ and f (x) → ∞ as x → + ∞. The degree is even and the leading coefficient is negative. So, f (x) → – ∞ as x → – ∞ and f (x) → – ∞ as x → + ∞. The degree is odd and the leading coefficient is positive. So, f (x) → – ∞ as x → – ∞ and f (x) → + ∞ as x → + ∞.

19. How will I EVER remember this?!? The degree of a line is 1 which is odd, the slope is positive. The degree of a line is 1 which is odd, the slope is negative. The degree of a parabola is 2 which is even, “a” is positive. The degree of a parabola is 2 which is even, “a” is negative. CV

20. Try this one… 8. Describe the degree and leading coefficient of the polynomial function whose graph is shown. ANSWER degree: odd, leading coefficient: negative

21. From the graph, f (x) → – ∞ as x → – ∞ and f (x) → – ∞ as x → + ∞. So, the degree is even and the leading coefficient is negative. The correct answer is D.ANSWER

22. Graphing Polynomial Functions Graph (a) f (x) = – x 3 + x 2 + 3x – 3 and (b) f (x) = 5 x 4 – x 3 – 4x 2 + 4. SOLUTION To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior. a. The degree is odd and leading coefficient is negative. So, f (x) → + ∞ as x → – ∞ and f (x) → – ∞ as x → + ∞.

23. Graphing Polynomial Functions Graph (b) f (x) = 5 x 4 – x 3 – 4x 2 + 4. To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior. b. The degree is even and leading coefficient is positive. So, f (x) → ∞ as x → – ∞ and f (x) → ∞ as x → + ∞.

24. Solve a multi-step problem The energy E (in foot-pounds) in each square foot of a wave is given by the model E = 0.0029s 4 where s is the wind speed (in knots). Graph the model. Use the graph to estimate the wind speed needed to generate a wave with 1000 foot-pounds of energy per square foot. Physical Science SOLUTION Make a table of values. The model only deals with positive values of s.

25. STEP 2Plot the points and connect them with a smooth curve. Because the leading coefficient is positive and the degree is even, the graph rises to the right. STEP 3 Examine the graph to see that s 24 when E = 1000.  The wind speed needed to generate the wave is about 24 knots. ANSWER

26. Graph the Polynomial Function 9. f (x) = x 4 + 6x 2 – 3 To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior. SOLUTION x – 2 –1 0 1 2 y 37 4 –3 4 37

27. Graph the Polynomial Function To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior. SOLUTION 10. f (x) = 2x 3 + x 2 + x – 1 x – 3 –2 –1 0 1 2 y 40 9 0 –1.7 –3 2

28. Graph the Polynomial Function 11. f (x) = 4 – 2x 3 To graph the function, make a table of values and plot the corresponding points. Connect the points with a smooth curve and check the end behavior. a. SOLUTION x – 2 –1 0 1 2 y 20 6 4 –12 2

29. WHAT IF? If wind speed is measured in miles per hour, the model in Example 6 becomes E = 0.0051s 4. Graph this model. What wind speed is needed to generate a wave with 2000 foot-pounds of energy per square foot? 12. ANSWER about 25 mi/h.

30. 2.2 Assignment Day 2 Page 99, 18-24 even, 28-36 all, 41- 47 odd