Chapter 2 Using lines to model data Finding equation of linear models Function notation/Making predictions Slope as rate of change.

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Chapter 2 Using lines to model data Finding equation of linear models Function notation/Making predictions Slope as rate of change

2.1 Using lines to model data Records were taken over a period of 5 years of the numbers of baby girls born in Linea Hospital. The data is shown in the chart below. YearBaby Girls

Scattergram A graph of plotted ordered pairs Should include: –scaling on both axes –labels of variables and scale units Number of girls born

Approximately Linearly Related Sketch a line that comes close to (or on) the data points There are multiple lines that will reasonably represent the data.

Definitions Approximately linearly related – a set of points in a scattergram of data that lie close to/or on a line Model – mathematical description of an authentic situation Linear model – linear function, or its graph, that describes the relationship between two quantities in an authentic situation.

Making Predictions with Linear Models Approximately how many babies will be born in 2010? –708 When were 500 babies born? –Sept 2006

When to Use a Linear Function to Represent Data

Scattergrams are used to determine if variables are approximately linearly related. Warning: Draw the line that comes close to all data points, not the greatest number of points

Intercepts of a model Let t be the years after 1970, let p be the polar bear population. Sketch a linear function to describe the relationship What does the p-intercept represent? –Population of 24,000 polar bears in 1970 When will the polar bears become extinct? –2015 Years Since p t

For a function with independent variable t : –interpolation : when part of the model used whose t-coordinates are between the t- coordinates of two data points –extrapolation: when part of the model used whose t-coordinates are not between the t- coordinates of any two data points more faith losing faith no faith model breakdown – when prediction doesn’t make sense or estimate is a bad approximation

Modifying a model In 2005, there were 6,000 recorded polar bears Modify to show the population leveling out at 8,000 polar bears. Modify to show polar bears becoming extinct.

Group Exploration p 62-63

Quiz Vocabulary Identify independent/dependent variable Find the equation of a line given a graph or graph the line give the equation Change an equation to slope intercept form Determine if lines are parallel, perpendicular or neither.

2.2 Finding equations of linear models (word problems) The number of Americans with health insurance has decreased approximate linearly from to 84.2% in 2006 to 76.3% in Let h be the percent of people with health insurance and t be the years since Find an equation of a linear model. Years since 2000 Pct with Health Ins. th

h = mt + b (6, 84.2), (9, 76.3) 76.3 – – 6 3 h = -2.64t + b (84.2) = -2.64(6) + b = b = b *h = -2.64t m = _____________=______≈

Finding equations of linear models (data tables) YearsER Patients

Years Since 2000 ER Patients (hundreds) (6,592) (1,395) (2,428) 197 5

197 ≈ p = 39.4t + b (395) = 39.4(1) + b = b = b p = 39.4t

So why not skip the Scattergram? 1. Used to determine Approx. Linearly Related. 2. If Approx. Linearly related, scattergrams allow us to choose 2 points to find an equation. 3. Determine if the model fits the data.

2.3 Function Notation and Making Predictions Sometimes it’s easier to name a function, instead of using equations, tables, graphs, etc. Generally, functions of linear equations are named f –y = f(x) –f being a different variable name for y ex. f(x) = 2x + 3 is the same as y = 2x + 3

Input/Output functions f(input) = output –f(x) = 2x + 3 –x = 4 –f(4) = 2(4) + 3 –f(4) = 11 Evaluate f(x) = 4x – 4 at 2. –f(2) = 4(2) – 4 –f(2) = 4 We can also use g to represent a function –g(x) = 4x – 4, (f, g, & h are the most common)

Evaluating Functions f(x) = 6x² - 2x + 5, Evaluate f(-2) f(-2) = 6(-2)² - 2(-2) + 5 f(-2) = 6(4) - 2(-2) + 5 f(-2) = f(-2) = 33 g(x) = -8x + 2 Evalutate g(a + 3) and g(a +h) g(a + 3) = -8(a + 3) + 2 g(a + 3) = -8a -8(3) + 2 g(a + 3) = -8a – g(a + 3) = -8a – 22 g(a + h) = -8(a + h) + 2 g(a + h) = -8a – 8h + 2

Using a table to find input/output xg(x) Find g(3) Input 3 gives output 2 Find x when g(x) = 7 Output 7 are given by inputs 2 and 4.

Using an equation to find an input f(x) = -4x² Find x when f(x) = = -4x² √ 4 = √ x² x = 2 or x = -2 inputs are 2 and -2 for the output of -16 **Careful when you are asked to find f(x) you are looking for the y value not the x value**

Using graphs to find input/outputs Find g(0), g(-5) Find x when g(x) = 6 g(x) = 0 (0,5) (-5,-1) (1,6) (-4,0)

Function notation in models Non-authentic situation dependent variable independent variable y = f (x) function name Authentic situation (model) –dependent variable = f(independent variable)

Making Predictions and Finding Intercepts Using Equations When making a prediction about the dependent variable, substitute a chosen value for the independent variable. Then solve for the dependent variable and vice versa. When looking for the intercept of the independent variable, substitute 0 for the dependent variable and solve for the independent variable and vice versa.

Finding equations of linear models (data tables) YearsER Patients

Find the intercept of the dependent variable p = 32.84(0) = = What does this represent? Number of patients in the year 2000 Find the intercept of the independent variable 0 = 32.84t – = 32.84t – = 32.84t ≈ tWhat does this represent? –The ER was founded in approx 1989

Examples p = 32.84t Predict the number of ER patients in 2009 p = 32.84(9) = = ≈ 657 patients in 2009 Predict the year there will be 1,000 patients 1,000 = 32.84t ,000 – = 32.84t – = 32.84t ≈ t≈2019

Finding the Domain and Range of a Model You can work a maximum 12 hour shift from 6am to 6pm M-F at the hospital. Let I = f(t), be your weekly income at $10 hour. f(t) = 10t Domain- least you can work is zero hours, most you can work is 5(12) = 60 hours. D = 0 ≤ t ≤ 60 Range- plug domain values into equation f(0) = 10(0) = 0, f(60) = 10(60) = 600 R = 0 ≤ I ≤ 600