EXAMPLE 1 Use x-intercepts to graph a polynomial function Graph the function f (x) = (x + 3)(x – 2)2. 1 6 SOLUTION STEP 1 Plot: the intercepts. Because –3 and 2 are zeros of f, plot (–3, 0) and (2, 0). STEP 2 Plot: points between and beyond the x-intercepts.

EXAMPLE 1 Use x-intercepts to graph a polynomial function STEP 3 1 6 Determine: end behavior. Because f has three factors of the form x – k and a constant factor of , it is a cubic function with a positive leading coefficient. So, f (x) → – ∞ as x → – ∞ and f (x) → + ∞ as x → + ∞. STEP 4 Draw the graph so that it passes through the plotted points and has the appropriate end behavior.

EXAMPLE 2 Find turning points Graph the function. Identify the x-intercepts and the points where the local maximums and local minimums occur. a. f (x) = x3 – 3x2 + 6 b. g (x) = x4 – 6x3 + 3x2 + 10x – 3

EXAMPLE 2 Find turning points SOLUTION a. f (x) = x3 – 3x2 + 6 a. Use a graphing calculator to graph the function. Notice that the graph of f has one x-intercept and two turning points. You can use the graphing calculator’s zero, maximum, and minimum features to approximate the coordinates of the points. The x-intercept of the graph is x  –1.20. The function has a local maximum at (0, 6) and a local minimum at (2, 2). ANSWER

EXAMPLE 2 Find turning points SOLUTION b. g (x) = x4 – 6x3 + 3x2 + 10x – 3 a. Use a graphing calculator to graph the function. Notice that the graph of g has four x-intercepts and three turning points. You can use the graphing calculator’s zero, maximum, and minimum features to approximate the coordinates of the points. The x-intercepts of the graph are x  –1.14, x  0.29, x  1.82, and x  5.03. The function has a local maximum at (1.11, 5.11) and local minimums at (–0.57, –6.51) and (3.96, – 43.04). ANSWER

EXAMPLE 3 Maximize a polynomial model You are making a rectangular box out of a 16-inch-by-20-inch piece of cardboard. The box will be formed by making the cuts shown in the diagram and folding up the sides. You want the box to have the greatest volume possible. Arts And Crafts • How long should you make the cuts? • What is the maximum volume? • What will the dimensions of the finished box be?

EXAMPLE 3 Maximize a polynomial model SOLUTION Write a verbal model for the volume. Then write a function.

Maximize a polynomial model EXAMPLE 3 Maximize a polynomial model = (320 – 72x + 4x2)x Multiply binomials. = 4x3 – 72x2 + 320x Write in standard form. To find the maximum volume, graph the volume function on a graphing calculator. Consider only the interval 0 < x < 8 because this describes the physical restrictions on the size of the flaps.

EXAMPLE 3 Maximize a polynomial model From the graph, you can see that the maximum volume is about 420 and occurs when x  2.94. You should make the cuts about 3 inches long.The maximum volume is about 420 cubic inches. The dimensions of the box with this volume will be about x = 3 inches by x = 10 inches by x = 14 inches. ANSWER

EXAMPLE 1 Write a cubic function Write the cubic function whose graph is shown. SOLUTION STEP 1 Use the three given x - intercepts to write the function in factored form. f (x) = a (x + 4)(x – 1)(x – 3) STEP 2 Find the value of a by substituting the coordinates of the fourth point.

EXAMPLE 1 Write a cubic function – 6 = a (0 + 4) (0 –1) (0 –3) – 6 = 12a – = a 2 1 2 1 The function is f (x) = (x + 4) (x – 1) (x – 3). ANSWER CHECK Check the end behavior of f. The degree of f is odd and a < 0. So f (x) + ∞ as x → – ∞ and f (x) → – ∞ as x → + ∞ which matches the graph.

EXAMPLE 2 Find finite differences The first five triangular numbers are shown below. A formula for the n the triangular number is f (n) = (n2 + n). Show that this function has constant second-order differences. 1 2

EXAMPLE 2 Find finite differences SOLUTION Write the first several triangular numbers. Find the first-order differences by subtracting consecutive triangular numbers. Then find the second-order differences by subtracting consecutive first-order differences.

EXAMPLE 2 Find finite differences Each second-order difference is 1, so the second-order differences are constant. ANSWER