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3.2 Polynomial Functions and Their Graphs

3.2 Polynomial Functions and Their Graphs. Definition of a Polynomial Function. Let n be a nonnegative integer and let a n , a n -1 ,…, a 2 , a 1 , a 0 , be real numbers with a n ¹ 0. The function defined by f ( x ) = a n x n + a n-1 x n -1 + … + a 2 x 2 + a 1 x + a 0

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3.2 Polynomial Functions and Their Graphs

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  1. 3.2 Polynomial Functions and Their Graphs

  2. Definition of a Polynomial Function Let n be a nonnegative integer and let an, an-1,…, a2, a1, a0, be real numbers with an¹ 0. The function defined by f (x)=anxn+ an-1xn-1+…+a2x2+ a1x + a0 is called a polynomial function of xof degree n. The number an, the coefficient of the variable to the highest power, is called the leading coefficient. Q:What do first, second, and third degree polynomial functions look like?

  3. 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 Q: What is an example of a function we have graphed that is not smooth? Not continuous?

  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 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 In this case, it is helpful to remember the slope of a line.

  5. 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 In this case, it is helpful to remember quadratic functions’ rules.

  6. Rises right y x Falls left Text Example Use the Leading Coefficient Test to determine the end behavior of the graph of 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. How would we find when the graph of the function would pass through the x- and y- axes? Q: What happens to the value of y as x gets very large? Very small?

  7. Text Example Find all zeros of f(x)=-x4+4x3- 4x2. (We know from the degree that we have a maximum of ______ zeros.) SolutionWe find the zeros of fby setting f(x)equal to 0. -x4+4x3- 4x2 = 0 We now have a polynomial equation. x4-4x3+ 4x2= 0 Multiply both sides by -1. (optional step) x2(x2- 4x+ 4) = 0 Factor out x2. x2(x- 2)2= 0 Factor completely. x2= 0 or (x- 2)2= 0 Set each factor equal to zero. x= 0 x= 2Solve for x.

  8. Multiplicity and x-Intercepts If r is a zero of even multiplicity, (x-r)n=0, where n is even, 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. Think: Stop and consider the degree, the max number of zeros, and the maximum number of turning points. When you have a multiplicity of say, 2, what happens?

  9. EX: 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 • Based on what we have discussed so far: • What is the degree of the polynomial? • Where do the left and right hand sides point? • Where does the graph pass through, and where does the graph touch and turn around? • Try to graph this, what other information would be helpful? • GRAPH IT!

  10. Graphing a Polynomial Function f (x)=anxn+ an-1xn-1+ an-2xn-2 + ¼ +a1x + a0 (an¹ 0) • Use the Leading Coefficient Test to determine the graph's end behavior. • Find x-intercepts by setting f (x) = 0 and solving the resulting polynomial equation. If there is an x-intercept at r as a result of (x-r)k in 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).

  11. Graphing a Polynomial Function f (x)=anxn+ an-1xn-1+ an-2xn-2 + ¼ +a1x + a0 (an¹ 0) • Use symmetry, if applicable, to help draw the graph: a.y-axis symmetry: f (-x) =f (x) (all x’s by themselves raised to even power or inside absolute value) b. Origin symmetry: f (-x) =-f (x). (all x’s by themselves raised to odd power, no constants) 5. Use the fact that the maximum number of turning points of the graph is n - 1 to check whether it is drawn correctly. Also, the maximum number of x intercepts (zeros of the function) will be the degree of the polynomial.

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

  13. Text Example cont. Graph: f(x)=x4-2x2+ 1. Solution Step 2 Find the x-intercepts (zeros of the function) by setting____ =0. x=x=Solve for x. Q: Before looking ahead, what does the even multiplicity do?

  14. 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 ____. 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:

  15. Text Example cont. y Rises right Rises left ? x -1 1 Graph: f(x)=x4-2x2+ 1. Solution Step 3 Find the y-intercept. Replace ________with 0. There is a y-intercept at _____, so the graph passes through ( , ). Think: Would it have been possible to have a negative y-intercept ?

  16. Replace xwith -x. Text Example cont. Graph: f(x)=x4-2x2+ 1. Solution Step 4 (OPTIONAL) 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 = f(x) Because f(-x)= f(x), the graph of f is symmetric with respect to the y-axis. The following figure shows the graph.

  17. y x Text Example cont. 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. Together p 323 # 62, then Quiz? P 323 #

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