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Precalculus – MAT 129. Instructor: Rachel Graham Location: BETTS Rm. 107 Time: 8 – 11:20 a.m. MWF. Chapter One. Functions and Their Graphs. 1.1 – Lines in the Plane. Slope Equation of a line Point-Slope form Slope-intercept form General form Parallel and Perpendicular Lines
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Precalculus – MAT 129 Instructor: Rachel Graham Location: BETTS Rm. 107 Time: 8 – 11:20 a.m. MWF
Chapter One Functions and Their Graphs
1.1 – Lines in the Plane • Slope • Equation of a line • Point-Slope form • Slope-intercept form • General form • Parallel and Perpendicular Lines • Applications
Slope • Formula: (y2-y1)/ (x2-x1) • Denoted: m • In words: Change in y over change in x
Slope Generalizations • Positive slope – rises (left to right) • Negative slope – falls (left to right) • Horizontal line – slope = 0 • Vertical line – slope = undefined
Example 1.1.1 Pg. 11 # 7 (Just find the slope) Find the slope of the line from the given points. Points = (0,-10) and (-4,0)
Solution - Ex. 1.1.1 Use slope formula!!! (0 + 10) / (-4 – 0) = 10/(-4) m = -5/2
Line Equations • Point-Slope form • Y-intercept form • General form
Point-Slope Form • Formula: (y-y1) = m(x-x1) • one point and the slope. OR • two points. • study tip on pg. 5
Y-Intercept Form • Most common • Formula: y = mx + b • Can be found by solving for y in the point-slope form or the general form. • Easy to sketch a line from this equation.
General Form • Formula: Ax + By = C • Can also be found from either of the other forms.
Example 2.1.1 Pg. 12 # 25 Find the general form of the line from the given point and slope. Point = (0,-2) Slope = 3
Solution - Ex. 2.1.1 Use the point-slope formula!!! (y + 2) = 3(x – 0) y + 2 = 3x -3x + y = -2
Example 3.1.1 Pg. 12 # 29 Find the general form of the line from the given point and slope. Point = (6,-1) Slope = undefined
Solution - Ex. 3.1.1 If the slope is undefined we know that it is a vertical line. A vertical line only crosses the x-axis. So all we need is the x-value from the point given. x = 6 This is also in general form.
Parallel and Perpendicular Lines • If two lines are parallel they have the same slope. • If two lines are perpendicular their slopes are negative reciprocals of each other. • “Change the sign and flip it!”
Example 4.1.1 Pg. 12 # 57 Write the slope-intercept form of the equation of the line through the give point a) parallel to the line given. b) perpendicular to the line given. Point = (2,1) Line => 4x -2y = 3
Solution - Ex. 4.1.1 Solving the given equation for y we get: y = 2x – 3/2 • same slope (use the point-slope form) (y – 1) = 2(x – 2) y – 1 = 2x – 4 y = 2x – 3
Solution - Ex. 4.1.1 b) negative reciprocal slope (use the point-slope form). (y – 1) = (-1/2)(x – 2) y – 1 =(-1/2x) + 1 y = (-1/2x) + 2
Example 5.1.1 Pg. 12 # 53 Determine whether the lines L1 and L2 passing through the pairs of points are parallel, perpendicular, or neither. L1: ( 0,-1) and (5,9) L2: (0, 3) and (4,1)
Solution - Ex. 5.1.1 L1: Slope = (9 + 1)/(5 – 0) = 2 L2: Slope = (1 – 3)/(4 – 0) = -2/4 = -1/2 m1 * m2 = -1 so these are negative reciprocals L1 and L2 are perpendicular.
Example 6.1.1 Pg. 13 # 71 Looking at the graph: a) Use the slopes to determine the year(s) in which the earnings per share of stock showed the greatest increase and decrease. b) Find the equation of the line between the years 1992 and 2002.
Example 6.1.1 Pg. 13 # 71 (cont.) c) Interpret the meaning of the slope from part (b) in the context of the problem. d) Use the equation from part (b) to estimate the earnings per share of stock for the year 2006. Do you think this is an accurate estimation? Explain.
Solution - Ex. 6.1.1 a) The greatest increase was between 1998 and 1999. b) 2 points (2, 0.58) and (12, 0.08) gives the slope -0.05. Use point-slope: (y – 0.58)=-0.05(x-2) y =- 0.05x + 0.68
Solution - Ex. 6.1.1 c) For every year increase there is a 0.05 decrease in earning per share. d) Plug in the appropriate number: y= -0.05(16) + 0.68 = -0.12 This is not accurate because our line does not accurately represent the data.
Example 7.1.1 Pg. 14 # 85 A controller purchases a bulldozer for $36,500. The bulldozer requires an average expenditure of $5.25/hr for fuel and maintenance, and the operator is paid $11.50/hr. a) Write a linear equation giving the total cost (C) of operating the bulldozer for t hours. (Include the purchase cost of the bulldozer)
Example 7.1.1 Pg. 14 # 85 (cont.) b) Assuming that customers are charged $27/hr of bulldozer use, write an equation for the revenue (R) derived from t hours of use. c) Use the profit formula (P=R-C) to write an equation for the profit derived from t hours of use.
Example 7.1.1 Pg. 14 # 85 (cont.) d) Use the result of part (c) to find the break-even point. (The number of hours the bulldozer must be used to yield a profit of $0.
Solution - Ex. 7.1.1 a) C = 16.75t + 36,500 b) R = 27t c) P = 10.25t – 36,500 d) t ≈ 3561 hours (graph and approx. where the lines cross)
1.2 – Functions • Definitions • Testing for functions • Evaluating a function • Domain of a function • Applications • Difference Quotients
Definitions • function –a function is a relationship between two variables such that to each value of the independent variable there corresponds exactly one value of the dependent variable • domain –the domain of a function is the set of all values of the independent variable for which the function is defined. • range – the range of a function is the set of all values assumed by the dependent variable
Definitions • X – independent variable • Y – dependent variable
Testing for Functions • If you are given the points check to see if there are any of the same x-values. • If so, then it is not a function. • Easiest way to test if a line is a function is to graph it and do the vertical line test. • Solve for y • Graph the line • Do the vertical line test (if only touches once then it is a function)
Evaluating a function • This is where we are putting something in our bucket (the variable). • At a given x-value what is the y-value?
Example 1.1.2 Pg. 19 Ex. 3 – Evaluating a Function Let g(x) = -x2 + 4x + 1. Find: (a) g(2) (b) g(t) (c) g(x+2).
Solution - Ex. 1.1.2 a. Replace x with 2 in g(x): g(2) = -(2)2 + 4(2) + 1 = 5 b. Replace x with t in g(x): g(t) = -(t)2 + 4(t) + 1 = -t2 + 4t + 1
Solution - Ex. 1.1.2 c. Replace x with (x + 2) in g(x): g(x + 2) = -(x + 2)2 + 4(x + 2) + 1 = -(x2 + 4x + 4) + 4x + 8 + 1 = -x2 + 5
Piecewise-Defined Function • A function that is defined by two or more equations over a specified domain. • See example pg. 19 in beige box.
Example 2.1.2 • On the board (19)
Domain of a function Domain is the set of all real numbers for which the expression is defined. • It’s as easy as traveling along the x-axis on the road that is your function. You can also figure out where the function cannot be defined.
Example 3.1.2 Pg. 26 #55 Find the domain of the function. h(t) = 4/t
Solution - Ex. 3.1.2 Pg. 26 #55 All real values of t except for t=0.
Applications • Go over Example 8 on pg. 22
Difference Quotients • Basic definition in calculus: ( f(x + h) – f(x)) / h, h ≠ 0
Example 4.1.2 Pg. 29 #89 Find the difference quotient and simplify your answer. f(x) = x2 – x + 1, (f(2 + h) – f(2))/h, h≠0. Work on the board!!
Activities (23) 1. Evaluate f(x) = 2 + 3x – x2 for: a. f(-3) b. f(x + 1) c. f(x + h) – f(x) 2. Find the domain: f(x) = 3/(x+1).
1.3 – Graphs of Functions • The Graph of a Function • Increasing and Decreasing Functions • Relative Minimum and Maximum Values • Graphing Step Functions and Piecewise-Defined Functions • Even and Odd Functions
The Graph of a Function • x = the directed distance from the y-axis. • y = the directed distance from the x-axis. • Go over Example 2 pg. 31 • Note both Algebraic and Graphical solutions.
Increasing and Decreasing Functions • Increasing <- the function is rising on the interval • Decreasing <- the function is falling on the interval • Constant
Example 1.1.3 Pg. 39 #21 Determine the intervals over which the function is increasing, decreasing, or constant. f(x) = x3 – 3x2 + 2 Graph on calculator. Draw on the board.