Analyzing Graphs of Functions Section P.6 Part 2.

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Presentation transcript:

Analyzing Graphs of Functions Section P.6 Part 2

Vertical Line Test (pencil test) If any vertical line passes through more than one point of the graph, then that relation is not a function. Are these functions? FUNCTION! FUNCTION! NOPE!

Vertical Line Test FUNCTION! NO! NO WAY! FUNCTION!

Is this a graph of a function? Yes No

A function is increasing on an interval if, for any x1 and x2 in the interval, x1 < x2 implies f (x1) < f (x2). Increasing, Decreasing, and Constant Functions Increasing f (x1) < f (x2) (x1, f (x1)) (x2, f (x2))

A function is decreasing on an interval if, for any x1 and x2 in the interval, x1 < x2 implies f (x1) > f (x2). Increasing, Decreasing, and Constant Functions Decreasing f (x1) > f (x2) (x1, f (x1)) (x2, f (x2))

A function is constant on an interval if, for any x1 and x2 in the interval, x1 < x2 implies f (x1) = f (x2). Increasing, Decreasing, and Constant Functions Constant f (x1) = f (x2) (x1, f (x1)) (x2, f (x2))

Keep the following things in mind: Read the graph like you would a book, from left to right. The behavior of the function is determined by the y-values, but the intervals are reported in terms of the x-values.

Describe the increasing, decreasing, or constant behavior The function is decreasing on the interval (-∞, 0), increasing on the interval (0, 2), and decreasing on the interval (2, ∞).

Describe the increasing, decreasing, or constant behavior The function is constant on the interval (-∞, 0). The function is increasing on the interval [0, ∞).

List the intervals on which the function is increasing, decreasing, and constant.

List the intervals on which the function is increasing, decreasing, and constant.

Maximum and Minimum Look where the function changes behavior. These points often represent minimum or maximum values of the function.

During the 1980s, the average price of a 1-carat polished diamond decreased and then increased according to the model: C = -0.7t³ + 16.25t² – 106t + 388, 2 ≤ t ≤ 10 C is the average price in dollars and t represents the year with t = 2 corresponding to 1982. The Price of Diamonds

The Price of Diamonds C = -0.7t³ + 16.25t² – 106t + 388, 2 ≤ t ≤ 10 According to this model, during which years was the price of diamonds decreasing and increasing? Approximate the minimum price of a 1-carat diamond between 1982 and 1990. The Price of Diamonds

First, sketch an accurate graph of this function: C = -0.7t³ + 16.25t² – 106t + 388, 2 ≤ t ≤ 10 First, sketch an accurate graph of this function: Solution 300 200 100 250 150 50 Price Per Carat (in dollars) Year ( 2  1982)

From the graph, you can see the price of diamonds decreased from 1982 to late 1984. From 1984 to 1990, the price increased. The minimum price was about $175. Solution 300 200 100 250 150 50 Price Per Carat (in dollars) Year ( 2  1982)

Classwork: Worksheet – Relations and Functions

Homework: Pg 78 – 82 Exercises: 1 – 12, 19 – 22 (Part a), 81