Download
1 / 29
290 likes | 336 Views
A program focusing on real numbers, exponents, algebra, rational expressions, polynomial equations, and linear functions. Learn to define, specify, and graph functions algebraically, numerically, and graphically. Master linear functions and solve systems of equations.
E N D
Elementary Mathematics Brush-Up
Why ? Situation: • Lot of different backgrounds in PUB, • Many of those bgs: non-mathematical… Observation: • People start doubting “obvious” things!
Program ? Focus is everything, so: • Not: real numbers, (rational) exponents, multiplying and factoring algebraic equations, rational expressions, solving polynomial equations, … (see 1st 15 pages of booklet) • Functions: numerical, algebraic & graphical viewpoint (with special attention to linear functions) • Systems of 2 equations in 2 unknowns: graphical & algebraic solution (elimination)
Functions Definitions: • A real-valued function f of a real variable is a rule that assigns to each real number x in a specified set of numbers, called the domain of f, a single real number f(x). • The variable x is called the independent variable. If y = f(x) we call y the dependent variable. • A function can be specified: - numerically - algebraically - graphically
Functions Numerically Specified Functions: • Suppose that the function f is specified by the following table: • Then, f(0) is the value of the function when x = 0. From the table, we obtain: • And so on…
Functions Algebraically Specified Functions: • Suppose that the function f is specified by f(x) = 3x2- 4x + 1. Then
Functions Graphically Specified Functions: • The graph of the function f is the set of all points (x, f(x)) in the xy-plane, where we restrict the values of x to lie in the domain of f. • To obtain the graph of a function, plot points of the form (x, f(x)) for several values of x in the domain of f. The shape of the entire graph can usually be inferred from sufficiently many points.
Functions Graphically Specified Functions: • Example: to sketch the graph of the functionf(x) = x2 with domain the set of all real numbers, first choose some values of x in the domain and compute the corresponding y-coordinates: • Plotting these points gives the picture on the left, suggesting the graph on the right:
Functions Linear Functions: • A linear function is one that can be written in the form: where m and b are fixed numbers. (the names m and b are traditional) • A linear function is one whose graph is a straight line (hence the term "linear"). • The graph of the above example looks like this:
Functions Linear Functions: y = mx + b • Example: y = 3x – 1 • Numerically, b is the value of y when x = 0 • Graphically, b is the y-intercept of the graph • Numerically, y increases by m units for every 1-unit increase of x • Geometrically, the graph rises by m units for every 1-unit move to the right; m is the slope of the line
Functions Linear Functions: y = mx + b Exercices: (1’) • y = 2x – 4 • y = x + 3 • y = -2x + 4 • 1 – y/6 = x/2 Graph ???
Functions Linear Functions: y = mx + b Finding the equation of a linear function, when given: • Slope & y-intercept immediately !!! • Example: Slope = 4 (= m!) y-Intercept = -3 (= b!) y = 4x – 3 !
Functions Linear Functions: y = mx + b Finding the equation of a linear function, when given: • Slope & y-intercept immediately !!! • Slope & Point (x1, y1) Slope-Intercept Formula: Direct Formula: y = y1 + m(x - x1) Example: (1,2) & slope 3 y = 2 + 3(x-1) = 2 + 3x – 3 = 3x – 1 y = mx + b, where b = y1 – mx1
Functions Linear Functions: y = mx + b Finding the equation of a linear function, when given: • Slope & y-intercept immediately !!! • Slope & Point (x1, y1) Slope-Intercept Formula Direct Formula • Points (x1, y1) & (x2, y2) calculate slope & intercept: m = (y2 – y1)/(x2 – x1) & b = y1 – mx1 (DF) Example: (0,1) & (2,-3) m = (-3 – 1)/(2 – 0) = -2 b = 1 – (-2)0 = -3 – (-2)2 = 1 y = -2x + 1
Functions Linear Functions: y = mx + b Exercises: (1’) • y-intercept = -4 & slope = 9 • Slope = 1/3 & point (0,-1) • Point (2,1) & slope = -1 • Points (-2,1) and (-1,2) Equation ???
Functions Linear Functions: y = mx + b Exercises: (1’) • y-intercept = -4 & slope = 9 y = 9x – 4 • Slope = 1/3 & point (0,-1) y = (1/3)x – 1 • Point (2,1) & slope = -1 y = -x + 3 • Points (-2,1) and (-1,2) y = x + 3
Systems of 2 Equations in 2 ?’s Definitions: • A linear equation in two unknowns x and y is an equation of the form ax + by = c, where a, b, and c are numbers, and where a and b are both not zero. • Examples: 4x + 5y = 0 This has a = 4, b = 5, c =0 x – y = 11 This has a = 1, b = -1, c = 11
Systems of 2 Equations in 2 ?’s Definitions: • A system of linear equations is just a collection of these beasts. To solve a system of linear equations means to find a solution (or solutions) (x, y) that simultaneously satisfies all of the equations in the system. • Examples: 4x + 5y = 40 x – y = 1 -x + 2y = 22 5x + 7y = 3 { {
Systems of 2 Equations in 2 ?’s Graphical Solution: • To locate solutions to a system of two equations in two unknowns, plot the graphs, and locate the intersection points (if any). • Examples: 2x + y = 4 2x – y = 2 solution = (1,5;1) {
Systems of 2 Equations in 2 ?’s Graphical Solution: • Examples: x – 2y = -2 x – 2y = 2 no solution ! (the lines are parallel) {
Systems of 2 Equations in 2 ?’s Graphical Solution: • Examples: x – 2y = -2 -2x + 4y = 4 infinitely many solutions ! (both equations are represented by the same line) General solution: x = 2y – 2 y is arbitrary { {
Systems of 2 Equations in 2 ?’s Algebraical Solution: • Problem with graphical approach: only gives approximate solutions! • Now: “method of elimination” = algebraic way of obtaining the exact solution(s) of a system of equations in two unknowns by manipulating the equations in such a way as to eliminate of the variables (x or y).
Systems of 2 Equations in 2 ?’s Algebraical Solution (“Elimination”): • Examples: 2x + 3y = 4 (1) 2x + 3y = 4 (1) x – 3y = 2 (2) 3x + 0y = 6 (1) + (2) 22 + 3y = 4 x = 2 y = 0 x = 2 { { { {
Systems of 2 Equations in 2 ?’s Algebraical Solution (“Elimination”): • Examples: 2x + 3y = 3 (1) 6x + 9y = 9 (1) 3 3x – 2y = -2 (2) -6x + 4y = 4 (2) (-2) 0x + 13y = 13 y = 1 6x + 9 1= 9 x = 0 y = 2 { { { {
Systems of 2 Equations in 2 ?’s Algebraical Solution (“Elimination”): • Examples: 3x – 5y = 1 (1) 6x – 10y = 2 (1) 2 -6x + 10y = -2 (2) -6x + 10y = -2 (2) 0x + 0y = 0 ?!? ! Infinitely many solutions! x = (5/3)y + (1/3) y is arbitrary { { { General solution:
Systems of 2 Equations in 2 ?’s Algebraical Solution (“Elimination”): • Examples: 3x – 5y = 1 (1) 6x – 10y = 2 (1) 2 -6x + 10y = 2 (2) -6x + 10y = 2 (2) 0x + 0y = 4 ?!? ! NO solutions ! (cf. graphically: 2 parallel lines no intersection…) { {
Systems of 2 Equations in 2 ?’s Algebraical Solution (“Elimination”): • Exercise: (1’) 4x – 3y = 2 (1) x + y = 1 (2) { …
Systems of 2 Equations in 2 ?’s Algebraical Solution (“Elimination”): • Exercise: (1’) 4x – 3y = 2 (1) 4x –3y = 2 (1) x + y = 1 (2) 3x + 3y = 3 (2) 3 7x + 0y = 5 x = 5/7 4(5/7) - 3y = 2 x = 5/7 x = 5/7 x = 5/7 (20/7) – (21/7)y = (14/7) 21y = 20 – 14 y = 6/21 { { { { { {
Thank you !!! For further assistance: Filip Goeman Operations & Technology Management Centre (1st floor) filip.goeman@vlerick.be
