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Algebra 2 Chapter 2 Notes Linear Equations and Functions PowerPoint Presentation

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Algebra 2 Chapter 2 Notes Unsubdivided Equations and Functions

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Algebra 2 Chapter 2 Notes Linear Equations and Functions

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  1. Algebra 2 Chapter 2 Notes Analogue Equations and Functions

  2. 2.1 Functions and their Graphs • A RELATION is a mapping or union of input values with output values. The set of input values in the DOMAIN and the set of production values is the Rove. A coitus is a FUNCTION if there is incisively one outturn for from each one input. It is not a function if at lest one input has more than one output. • ( x , y ) = (domain , range ) = ( input signal , output) = (independent , dependent) • Not a functionYes a role • 3 3 - 3 3 • 1 –2 1 1 • 1 2 • 4 4 4 –2 • A relation is a function if and only when atomic number 102 straight-backed line intersects the graphical record of the relationship at more than cardinal point.         Non a RELATION YES a RELATION

  3. 2.1 Coordinate Plane Y axis vertebra Quadrant II ( – , + ) Quadrant I ( + , + ) ( 0 , 0 ) X axis Quadrant Trey ( – , – ) Quadrant Foursome ( + ,– ) x coordinate is first y coordinate is irregular Ordered pairs in constitute of ( x , y )

  4. 2.1 Functions and their Graphs y = m x + b LINEAR Run f (x) = m x + b FUNCTION Annotation • Graphing equations in 2 variables • Fabricate a table of values • Graph enough solutions to agnise a pattern • Unite the points with a furrow or a curve ball y Graph the officiate: y = x + 1     x 

  5. 2.1 Functions and their Graphs • Are these functions one-dimensional?. Evaluate when x = – 2 • f ( x ) = – x2 – 3 x + 5 Non a work because x is to the 2nd power • f ( – 2 ) = – ( – 2 ) 2 – 3 ( – 2 ) + 5 • f ( – 2 ) = 7 • g ( x ) = 2 x + 6 Yes a function because x is to the 1st power • g ( – 2 ) = 2 ( – 2 ) + 6 • g (– 2 ) = 2

  6. 2.1 Pitch and Rank of Change SLOPEof a non-unbowed subscriber line is the ratio of a vertical change (RISE) to a horizontal change (RUN). Slope of a line: m = y2 – y1 = RISE x2 – x1= RUN (differences in y values) (differences in x values) y ( x2,y2) •  y2 – y1 RISE ( x1,y1) • x2 – x1 RUN x

  7. 2.2 Slope and Rank of Change CLASSIFICATION OF LINES Past SLOPE A describe with a +slope rises from left-hand to right [ m > 0 ] Positive Slope A wrinkle with a –slope waterfall from left to right [ m < 0 ] Harmful Slope A line with a slope of 0 is horizontal [ m = 0 ] 0 Slope A line with an undefined slope is vertical [ m = undefined, no slope ] None Slope

  8. 2.2 Pitch and Rate of Change SLOPES OF PARALLEL & PERPENDICULAR LINES PARALLEL LINES: the lines are parallel if and only they have the SAME Gradient. M1 = M2 PERPENDICULAR LINES: the lines are perpendicular if and only if their SLOPES are NEGATIVE RECIPROCALS. m1 = –1 M2 operating theater m1m2 = – 1

  9. 2.2 Slope and Rate of Variety Gradient of a line: m = y2 – y1 = RISE (differences in y values) x2 – x1= RUN (differences in x values) Demode 1: Uncovering the slope of a line passing done ( – 3, 5 ) and ( 2, 1 ) Let ( x1, y1 ) = ( – 3, 5 ) and ( x2, y2 ) = ( 2, 1 ) m = 5 – 1 = 4 – 3 – 2 – 5 Operating theater m = 1 – 5 = – 4 2 + 3 5 5 • ( – 3, 5 ) –4 • x ( 2, 1 ) y

  10. 2.2 Incline and Order of Change • Example 2: Without graphing tells if slope rises, falls, horizontal or unsloped: • ( 3, – 4 ) and ( 1, – 6 ) : m = – 6 – ( – 4 ) = - 2 = 1 m > 0, rises • 1 – 3 - 2 • b) ( 2, 2 ) and ( –1, 5 ) : m = 5 – ( – 1 ) = 6 = undefined m = no slope • 2 – 2 0 For 2 lines with + slopes, the line with > slope is steeper For 2 lines with – slopes, the line with slope of > living value is steeper: y m = 3 m = - 3 m = 1 m = - 1 m = 1/2 m = - 1/2 x

  11. 2.2 Classifying Lines Using Slopes Classifying Vertical Lines Blood line 1 through ( – 3 , 3) and ( 3 , – 1 ) m 1 = – 1 – 3 = – 4 = – 2 3 – ( - 3) 6 3 Line 2 through ( - 2 , - 3 ) and ( 2 , 3 ) y L2 L1 ( –3 , 3 ) • • ( 2, 3 ) x • m 2 = 3 – ( – 3 ) = 6 = 3 2 – ( – 2 ) 4 2 ( 3 , – 1 ) Because M1 ∙ m2 = – 2 ∙ 3 = – 1 3 2 are negative reciprocals of each other, the lines are perpendicular. • ( –2 , – 3 )

  12. 2.2 Classifying Lines Using Slopes Classifying Parallel Lines Line 1 through ( – 3 , 3) and ( 3 , – 1 ) m 1 = 4 – 1 = 3 = 1 3 – ( - 3) 6 2 Line 2 through ( - 2 , - 3 ) and ( 2 , 3 ) ( 3, 4 ) y • ( 4 , 1 ) ( -3 , 1 ) • • x m 2 = 1 – ( – 3 ) = 4 = 1 4 – ( – 4 ) 8 2 L1 ( -4 , - 3 ) Because m1 = m2 and the lines are different, then the lines are parallel. • L2

  13. 2.3 Quick Graphs of Rectilineal Equations Slope Stop Form of a linear equation is y = m x + b, where m is side and b is y-intercept • Graphing Equations in side-intercept form: • Write the equation in slope-intercept shape past solving for y • Notic y-tap, then plot the channelis where line crosses the y-axis • Obtain the incline and use IT to plot a second point on the line. • Draw a line finished the 2 points. y • Example 1: Graphing with the slope-intercept • Graph: y = 3 x − 2 • 4 • Already in slope-intercept descriptor • y -stop is – 2, plot point ( 0 , – 2 ) • where the line crosses the y-axis • 3. Side is ¾ , so plot 4 units to right, • 3 units finished, steer is ( 4 , 1 ) • 4. Pass over a line through the 2 points ( 4 , 1) • x 3 • ( 0 , – 2) 4

  14. 2.3 Quick Graphs of One-dimensional Equations Standard Form of a linear equating is ax + by = c x-intercept of a line is the x-coordinate of the point where the line intersects the x-axis • Graphing Equations in common mold • Write out the equation in standard form • Find x-tap, by letting y = 0, work out for x, plot the point where x crosses the x-axis • Find y-intercept, by letting x = 0, solve for y, game the taper off where y crosses the y-axis of rotation • Draw a note direct the 2 points. • Method 1 by using Definitive Form: • Graphical record 2 x + 3 y = 12 • Already exploitation standard form: 2 x • Let y = 0 2 x + 3(0) = 12 • 2 x = 12 • x = 6 • x-intercept is at ( 6 , 0 ) • Let x = 0 2 (0) + 3 y = 12 • 3 y = 12 • y = 4 • y-stop is at ( 0 , 4 ) • 2 x + 3 y = 12 ( 0 , 4 ) • ( 6 , 0 )

  15. 2.3 Quick Graphs of Linear Equations • 3 • Method 2 by using Slope-Intercept Form: • Graph 2 x + 3 y = 12 • Change from stock form to slope-intercept form: 2 x + 3 y = 12 • -2 x - 2 x • 3 y = – 2 x + 12 • 3 3 • y = – 2 x + 4 • 3 • Identify , plot of ground y-intercept • Practice slope to patch early charge • Draw the line finished the points ( 0 , 4 ) 2 • ( 3 , 2 ) 2 x + 3 y = 12 y-intercept incline y x = 3 ( 0 , 3 ) y = 3 • Crosswise lines --- graph of y = c is a horizontal line through and through, (0 , c) Vertical lines --- graph of x = c is a vertical line through (c, 0) Graph y = 3 and x = -2 ( - 2 , 0 ) • x

  16. 2.4 Writing Equations of Lines Slope-Intercept Form: y = m x + b Point-Intercept Form: y – y1 = m ( x – x1 ) Two Points: m = y2 – y1 x2 – x1 * Every non-vertical bloodline has only one slope and one y-intercept m = slope b = y-intercept y • Write an equation of line shown • From graph you can escort the slope, m = 3 • 2 • From graph you can see that IT intersects at ( 0 , −1 ) , so the y-intercepts is b = −1 • ( 2, 2) x 3 • Equation of the Line is: y = 3 x − 1 2 ( 0, −1) 2

  17. 2.4 m ( x1 ,y1 ) Writing equations of the line given slope of − 1 and a channelize, ( 2 , 3 ) 2 Writing Equations of Lines y – y1= m ( x – x1) y – 3= –1( x – 2) 2 y – 3= –1x + 1 2 + 3 + 3 y = – 1 x + 4 2 Equivalence of the Line

  18. 2.4 Writing Equations of Orthogonal and Parallel Lines y – y1 = M2 ( x – x1) y – 2 = 1 ( x – 3) 3 y – 2 = 1 x – 1 3 + 2 + 2 y = 1 x + 1 3 Exemplar 3a: Equation of a line that passes through ( 3 , 2 ) that is perpendicular to the line: y = – 3 x + 2 Perpendicular means m2 = – 1 m1 Example 3b: Equation of a line that passes direct ( 3 , 2 ) that is parallel to the line: y = – 3 x + 2 Parallel means M2 = M1 = – 3 and ( x1 , y1 ) = ( 3 , 2 ) y – y1 = m2 ( x – x1) y – 2 = – 3 ( x – 3) y – 2 = – 3 x + 9 + 2 + 2 y = – 3 x + 11

  19. 2.4 Written material Equations of Lines Writing Equations given 2 points where ( x1 , y1 ) = ( – 2 , –1 ) ( x2 , y2 ) = ( 3 , 4 ) m = y2 – y1 = 4 – ( – 1 )= 5 = 1 x2 – x1 3 – ( – 2 ) 5 y – y1 = m ( x – x1) y – ( – 1 ) = 1 [ x – ( – 2 ) ] y + 1 = 1 ( x + 2 ) y = x + 1

  20. 2.4 Writing the Equation of the Communication channel The form you use depends on the info you have

  21. 2.5 Writing Pointed Variation Equations • Man-to-man Variation is shown away y = k x , where k ≠ 0 and k is a constant, (that is a single valuate) • Example 1: Variables x and y vary directly, y = 12 and x = 4 • write and graph an par for y and x • find y when x = 5 • y x y 0 0 4 12 5 15 • y = kx (12 ) = k (4) 3 = k • x y = 3x Example 3 x = 9 and y = 15 y = kx 15 = k9 5= k 3 Exercise 2 x = 6 and y = 3 y = kx 3 = k6 1= k 2 y = 1x 2 y = 5x 3

  22. 2.5 Scatter Plots A scatter plot is a graph used to determine, whether there is a relationship between paired information. Positive Correlation Negative Correlation • • • • • • • • • • • • • • • • • • • • • • • • No Correlation • • • • • • • • • • • • • • • •

  23. 2.5 A scatter plot is a graphical record used to determine, whether on that point is a family relationship between paired information Approximating a optimal-meet line: In writing approach Step 1: draw a scatter plot of dataStep 2: survey the line supported the pattern Step 3: choose 2 points on the personal line of credit Footfall 4: find the equation of the line that passes through the 2 points Example 2: Fitting a line to data Line through with two points: m = 1 − .6 = .4 = .25 2.5 - .9 1.6 Point-incline form: y – y1 = m ( x – x1) y – .6 = .25 ( x – .91) y – .6 = .25 x - .225 + .6 + .6 y = .25 x + .225 Scatter Plots • • 1.2 .8 .4 .2 • • • • • • • • • • • • 0 1 2 3

  24. 2.6 Linear Inequalities in Two Variables Linear Inequalities in Cardinal Variables can be handwritten in one of the following forms: ax + aside < c ax + by ≤ c ax + past > c axe + past ≥ c An coherent pair ( x , y ) is a answer of a simple inequality if the inequality is apodeictic for values of x , y. Example, ( − 6 , 2 ) is a solution of y ≥ 3 x − 9 because 2 ≥ 3 (− 6) − 9 is true as 2 ≥ − 27 Example 1: Checking Solutions of Inequalities Set pairs Substitute Stopping point ( 0 , 1 ) 2 ( 0 ) + 3 ( 1 ) ≥ 5 3 ≥ 5 False, non a solution ( 4 , −1 ) 2 (4 ) + 3 (−1 ) ≥ 5 5 ≥ 5 True, is a solution ( 2 , 1 ) 2 ( 2 ) + 3 ( 1 ) ≥ 5 7 ≥ 5 True, is a solution

  25. 2.6 Linear Inequalities in Two Variables Graph of a linear inequality in two variables is a half-plane. To graph a linear inequality adopt these steps: Step 1: Chart the mete of the inequality. Remember to practice a dash line for < or > and a unbroken line for ≤ or ≥ Step 2: Test a point out that is NOT on the line to see to it what side to shade. Graph: y < − 2 Graph: x < 1 y y x < 1 x = 1 x x y = − 2 y < − 2

  26. 2.7 Graphing One-dimensional Inequalities in Two Variables y = 2 x • • • Test Point ( 1, 1 ) y < 2 x Test Point ( 0, 0 ) 2 x – 5 y = 10 • • • 2 x – 5 y > 10

  27. 2.7 Piecewise Functions Piecewise Functions: A combination of equations, each corresponding to divide of a domain Example 1: Evaluating a piecewise work Evaluate f (x) when x = 0, when x = 2 and when x = 4 Evaluate f (x) = x + 2 , if x < 2 2 x + 1 , if x ≥ 2 Solution: When x = 0 , f (x) = x + 2 f (0) = 0 + 2 = 2 When x = 2 , f (x) = 2 (x) + 1 f (2) = 2 (2) + 1 = 5 When x = 4 , f (x) = 2 (x) + 1 f (4) = 2 (4) + 1 = 9 { • ○ 0 2

  28. 2.7 Graphing Piecewise Functions { Graphing a Piecewise Functions: f (x) = 1x + 3 , if x < 1 2 2 − x + 3 , if x ≥ 1 ( 1 , 2 ) • • • f (x) = 1 x + 3 2 2 • f (1) = 1 (1 ) + 3 = 4 = 2 2 2 2 f ( −3) = 1 ( −3) + 3 = 0 2 2 f ( x ) = − x + 3 f ( 1 ) = − ( 1 ) + 3= 2 f (3 ) = − (3) + 3 = 0 2 rays with a common first point, ( 1, 2 )

  29. 2.7 Graphing Piecewise Functions y Chart the following piecewise run, show all work, pronounce graph • { f (x) = 2x – 1 , if x ≤ 1 3 x + 1 , if x > 1 ○ • x •

  30. 2.7 Graphing Piecewise Functions { y ○ • 4 f (x) = 1 , if 0 ≤ x < 1 2 , if 1 ≤ x < 2 3 , if 2 ≤ x < 3 4 , if 3 ≤ x < 4 • ○ 3 • ○ 2 • ○ 1 x 1 2 3 4 It is known as a Step work because its graph looks like a set of stair steps

  31. 2.7 Writing a Piecewise Function (−2 , 0 ) , ( 0 , 2 ) m = 0 – 2 = − 2 = 1 − 2 – 0 − 2 ( 0 , 2 ) ○ • ( 2 , 2 ) y = 1 x + 2 (−2 , 0 ) • • ( 0 , 0 ) y = 1 x + 0 (0, 0 ) , ( 2 , 2 ) m = 0 – 2 = − 2 = 1 0 – 2 − 2

  32. y Writing a Piecewise Subprogram y = x y = − x { • • x , if x > 0 Remember: │x│ =0, if x = 0 − x, if x < 0 ( − 2 , 2 ) ( 2 , 2 ) x • y = │x │ Vertex Slope – Wiretap Manikin y = m x + b Absolute Value Function: y = a │ x – h │ + k

  33. 2.8 Absolute Value Functions Absolute Value Function: the a measure gives you face astir or down and steepness y = a │ x – h │ + k If a is + , then the graphical record is confront or If a is − , then the graph is confront down If a is bigger, then slope is steeper If a is little, then pitch is wider

  34. 2.8 Absolute Value Functions Absolute Value Function: the h apprais gives you Right (+) or unexpended (−) movement of vertex y = a │ x – h │ + k The more h is + , the more the vertex moves to the right The more h is − , the Sir Thomas More the vertex moves to the left y = a │ x – ( − 3) │ + k y = a │ x – ( 0 ) │ + k y = a │ x – ( 3 ) │ + k

  35. 2.8 Arbitrary Value Functions Absolute Value Function: the h value besides gives you Line of Proportion y = a │ x – h │ + k The line of proportion is in the line x = h y = a │ x – ( − 3) │ + k y = a │ x – ( 0 ) │ + k y = a │ x – ( 3 ) │ + k x = − 3 x = 0 x = − 3

  36. 2.8 Absolute Value Functions Absolute Value Function: the k value gives you Improving or retired movement of the apex y = a │ x – h │ + k The more k is + , the more the vertex moves risen The more k is − , the more the vertex moves Down y = a │ x – h │ + 3 y = a │ x – h │ + 0 y = a │ x – h│ − 3

  37. 2.8 Absolute Economic value Functions (h, k) values gives you the Vertex of the Absolute Treasure Function y = a │ x – h │ + k • (h, k)

algebra 2 chapter 2 linear relations and functions answers

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