4.2 Polynomial Functions and Models

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Presentation transcript:

4.2 Polynomial Functions and Models Understand the graphs of polynomial functions Evaluate and graph piecewise-defined functions Use polynomial regression to model data

Graphs of Polynomial Functions (1 of 3)

Graphs of Polynomial Functions (2 of 3) 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)

Graphs of Polynomial Functions (3 of 3)

Constant Polynomial Function If f(x) = a and a ≠ 0, then f is both a constant function and a polynomial function of degree 0. Graph is a horizontal line. No x-intercepts or turning points

Linear Polynomial Function (1 of 2) If f(x) = ax + b and a ≠ 0, then ƒ is both a linear function and a polynomial function of degree 1. Graph is a line, neither horizontal nor vertical. Graph has one x-intercept and no turning points

Linear Polynomial Function (2 of 2)

Quadratic Polynomial Functions (1 of 2) If f(x) = ax² + bx + c and a ≠ 0, then ƒ is both a quadratic function and a polynomial function of degree 2. Graph is a parabola, opening either upward (a > 0) or downward (a < 0). Graph can have zero, one, or two x-intercepts, and exactly one turning point, called the vertex.

Quadratic Polynomial Functions (2 of 2)

Cubic Polynomial Functions (1 of 2) If f(x) = ax³ + bx² + cx + d and a ≠ 0, then f is both a cubic function and a polynomial function of degree 3.

Cubic Polynomial Functions (2 of 2)

Quartic Polynomial Functions (1 of 2) Graph can have up to four x-intercepts and up to three turning points.

Quartic Polynomial Functions (2 of 2)

Quintic Polynomial Functions (1 of 2)

Quintic Polynomial Functions (2 of 2)

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.

Example: Analyzing the graph of a Polynomial function (1 of 8)

Example: Analyzing the graph of a Polynomial function (2 of 8) Use the graph of the polynomial function f shown. a. How many turning points and x-intercepts are there? b. Is the leading coefficient a positive or negative? Is the degree odd or even? c. Determine the minimum degree of f.

Example: Analyzing the graph of a Polynomial function (3 of 8) Solution a. There are four turning points corresponding to the two “hills” and two “valleys”. There appear to be 4 x-intercepts.

Example: Analyzing the graph of a Polynomial function (4 of 8) b. The left side rises and the right side falls. Therefore, a < 0 and the polynomial function has odd degree. c. The graph has four turning points. A polynomial of degree n can have at most n − 1 turning points. Therefore, f must be at least degree 5.

Example: Analyzing the graph of a Polynomial function (5 of 8) 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 (6 of 8) 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, 0), (1, 0), and (3, 0).

Example: Analyzing the graph of a Polynomial function (7 of 8) b. There are two turning points. Their coordinates approximately (−0.79, 8.21) and (2.12, −4.06).

Example: Analyzing the graph of a Polynomial function (8 of 8) c. There is a local maximum of about 8.21 and a local minimum of about −4.06.

Example: Analyzing the end behavior of a graph (1 of 2) 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 (2 of 2) b. The degree is odd and the leading coefficient is negative. The graph of f rises to the left and falls to the right. More formally,

Example: Evaluating a piecewise-defined Polynomial function (1 of 2) Evaluate f(x) at −3, −2, 1, and 2. Solution To evaluate f(−3) we use the formula x² − x because −3 is the interval −5 ≤ x < −2. f(−3) = (−3)² − (−3) = 12

Example: Evaluating a piecewise-defined Polynomial function (2 of 2) To evaluate f(−2) we use f(x) = −x³ because −2 is in the interval −2 ≤ x < 2. f(−2) = −(−2)³ = −(−8) = 8 Similarly, f(1) = −1³ = −1 and f(2) = 4 − 4(2) = −4

Example: Graphing a piecewise-defined function Complete the following. a. Sketch a graph of f. b. Determine if f is continuous on its domain. c. Solve the equation f(x) = 1.

Example: Graphing a piecewise-defined Polynomial function (1 of 2)

Example: Graphing a piecewise-defined Polynomial function (2 of 2) b. The domain is −4 ≤ x ≤ 4. Because there are no breaks in the graph of f on its domain, the graph of f is continuous.

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 previous sections, for linear and quadratic functions.

Example: Determining a cubic modeling function (1 of 5) The data in the table lists Brazil’s unemployment rates. Year Rate (%) 2010 6.75 2012 5.48 2014 4.84 2016 9.19 2018 10.4 2020 10.0

Example: Determining a cubic modeling function (2 of 5) a. Find a polynomial function of degree 3 that models the data. b. Graph f and the data together. c. Estimate unemployment in 2024. d. Did your estimates in part c involve interpolation or extrapolation? Is there a problem with using higher degree polynomials (n ≥ 3) for extrapolation? Explain.

Example: Determining a cubic modeling function (3 of 5) Solution a. Enter the six data points. Then select cubic regression. The equation for ƒ(x) is shown.

Example: Determining a cubic modeling function (4 of 5) b. Here is the scatterplot.

Example: Determining a cubic modeling function (5 of 5) c. f(2024) ≈ −6.58, which is incorrect because the employment rate cannot be negative.