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Chapter 4 More Nonlinear Functions and Equations. Polynomial Functions and Models. 4.2. Understand the graphs of polynomial functions Evaluate and graph piecewise-defined functions Use polynomial regression to model data (optional). Graphs of Polynomial Functions.
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Chapter 4 More Nonlinear Functions and Equations
Polynomial Functions and Models 4.2 Understand the graphs of polynomial functions Evaluate and graph piecewise-defined functions Use polynomial regression to model data (optional)
Graphs of Polynomial Functions A polynomial function f of degree n can be expressed as f(x) = anxn + … + a2x2 + a1x + a0, where each coefficient ak is a real number, an 0, and n is a nonnegative integer.
Graphs of Polynomial Functions A turning point occurs whenever the graph of a polynomial function changes from increasing to decreasing or from decreasing to increasing. Turning points are associated with “hills” or “valleys” on a graph: (–2, 8) and (2, –8)
Constant Polynomial Function f(x) = a Has no x-intercepts or turning points
Linear Polynomial Function f(x) = ax + b Degree 1 and one x-intercept and no turning points.
Quadratic Polynomial Functions f(x) = ax2 + bx + c Degree 2, parabola that opens up or down. Can have zero, one or two x-intercepts. Has exactly one turning point, which is also the vertex.
Cubic Polynomial Functions f(x) = ax3 + bx2+ cx + d Degree 3, can have up to 3 x-intercepts and zero or two turning points.
Quartic Polynomial Functions f(x) = ax4 + bx3+ cx2+ dx + e Degree 4, can have up to four x-intercepts and three turning points.
Quintic Polynomial Functions f(x) = ax5 + bx4+ cx3+ dx2+ ex + k Degree 5, may have up to five x-intercepts and four turning points.
Degree, x-intercepts, and Turning Points The graph of a polynomial function of degree n, with n ≥ 1, has at most n x-intercepts and at most n – 1 turning points.
Use the graph of the polynomial function f shown. a) How many turning points and x-intercepts are there? b) Is the leading coefficient Example: Analyzing the graph of a polynomial function a positive or negative? Is the degree odd or even? c) Determine the minimum degree of f.
Solution a) There are four turning points corresponding to the one “hill” and two “valleys”. Example: Analyzing the graph of a polynomial function There appear to be 4 x-intercepts.
b) The left side rises and the right side falls. Therefore, a < 0 and the polynomial function has odd degree. Example: Analyzing the graph of a polynomial function c) The graph has four turning points. A polynomial of degree n can have at mostn 1 turning points. Therefore, f must be at least degree 5.
Graph f(x) = x3 2x2 5x + 6, and then complete the following. a) Identify the x-intercepts. b) Approximate the coordinates of any turning points to the nearest hundredth. c) Use the turning points to approximate any local extrema. Example: Analyzing the graph of a polynomial function
Solution a) The graph appears to intersect the x-axis at the points (–2,0), (1, 0), and (3, 0). The x-intercepts are –2, 1, and 3. Example: Analyzing the graph of a polynomial function
b) There are two turning points. One has coordinates approximately (–0.79, 8.21). Example: Analyzing the graph of a polynomial function
b) The other has coordinates approximately (2.12, –4.06). Example: Analyzing the graph of a polynomial function
c) There is a local maximum of about 8.21 and a local minimum of about –4.06. Example: Analyzing the graph of a polynomial function
Let f(x) = 2 + 3x – 3x2 2x3. a) Give the degree and leading coefficient. b) State the end behavior of the graph of f. Solution a) The term with the highest degree is –2x3 so the degree is 3 and the leading coefficient is –2. Example: Analyzing the end behavior of a graph
b) The degree is odd and the leading Example: Analyzing the end behavior of a graph coefficient is negative. The graph of f rises to the left and falls to the right. More formally,
Evaluate f(x) at 3, –2, 1, and 2. Solution To evaluate f(3) we use the formula x2 – xbecause 3 is the interval 5 ≤ x < –2. f(3) = (3)2 – (3) = 12 Example: Evaluating a piecewise-defined polynomial function
To evaluate f(2) we use the formula x3because 2 is in the interval –2 ≤ x < 2. f(2) = (2)3 = –(–8) = 8 Similarly, f(1) = –13 = –1 To evaluate f(2) we use the formula 4 4xbecause 2 is in the interval 2 ≤ x ≤ 5. f(2) = 4 4(2) = 4 8 = 4 Example: Evaluating a piecewise-defined polynomial function
Complete the following. a) Sketch the graph of f. b) Determine if f is continuous on its domain. c) Find x-coordinates such that f(x)=1. Example: Graphing a piecewise-defined polynomial function
Solution a) Sketch the graph of f. Graph on the interval –4 ≤ x ≤ 0 Graph y = 2x – 2 between the points (0, –2) and (2, 2). Graph y = 2 from the points (2, 2) to (4, 2). Example: Graphing a piecewise-defined polynomial function
b) The domain is –4 ≤ x ≤ 4 c) The x-coordinates for the points of Example: Graphing a piecewise-defined polynomial function intersection can be found by solving2x – 2 = 1 and Solutions are:
Polynomial Regression We now have the mathematical understanding to model the data presented in the introduction to this section: polynomial modeling. We can use least-squares regression, which was also discussed in Sections 2.1 and 3.1, for linear and quadratic functions.
The table lists natural gas consumption. Example: Determining a cubic modeling function
(a) Find a polynomial function of degree 3 that models the data. (b) Graph f and the data together. (c) Estimate natural gas consumption in 1974 and in 2010. Compare these estimates to the actual values of 21.2 and 24.9 quadrillion Btu, respectively. (d) Did your estimates in part (c) involve interpolation or extrapolation? Is there a problem with using higher degree polynomials for extrapolation? Explain. Example: Determining a cubic modeling function
Solution a) Enter the data and select CubicReg. Example: Determining a cubic modeling function
a) Enter the 5 data points into your calculator. Then select cubic regression. You’ll get this display. Example: Determining a cubic modeling function
b) Here is the scatterplot. Example: Determining a cubic modeling function
c) f(1974) ≈ 21.9 and f(2010) 44.5; the 1974 estimate is reasonably close to 21.2, where as the 2010 estimate is not close to 24.9. d) The 1974 estimate uses interpolation, and the 2010 estimate uses extrapolation. Because of the end behavior tending towards ∞ and –∞, extrapolation-based estimates are usually inaccurate with higher degree polynomials. Example: Determining a cubic modeling function