Chapter 01 Linear and Quadratic Functions Notes …

1 PreCalculus Chapter 1: Linear and Quadratic Functions Chapter 1: Linear and Quadratic Functions 1-1: Points and Lines System of Linear Equations: - two or more Linear equations on the same coordinate grid. Solution of a System of Linear Equations: - the intersecting point of two or more Linear equations - on the Cartesian Coordinate Grid, the solution contains two parts: the x-coordinate and the y-coordinate Line 1 Line 2 (can be expressed as an Ordered Pair). 9. 8. 7. Solution of the System of Linear Equations 6. 5. 4. 3. 2 Intersecting Point (x, y). 1. 0. -9 -8 -7 -6 -5 -4 -3 -2 -1. -1 0 1 2 3 4 5 6 7 8 9. -2. -3. -4. -5. -6. -7 We can find the solution of a system of Linear equation -8 graphing manually or using a graphing calculator. -9. Example 1: Find the solution of the following system of equations by graphing manually using x + 2y = 6. x y =3. 9. a. the slope and y-intercept form 8. Line 1 7 Line 2. For Line 1: For Line 2: 6. 5. x + 2y = 6 x y =3 4.

2 2 y = x + 6 y = x + 3 (0, 3). 3 Intersecting 2. x+6 Point (4, 1). y= y = x 3 1 (2, 2). 2 0. y-int = (0, 3) -9 -8 -7 -6 -5 -4 -3 -2 -1. -1 0 1 2 3 4 5 6 7 8 9. 1 1 1 Up -2 (1, 2). y = x+3 slope = = -3. 2 1 1 Right (0, 3)-4. y-int = (0, 3) -5. 1 1 Down -6. slope = =. 2 2 Right -7. -8. -9. Copyrighted by Gabriel Tang , Page 1. Chapter 1: Linear and Quadratic Functions PreCalculus b. x and y-intercepts 9. Line 2. Line 1 8. Intersecting Line 1: Line 2: 7. 6. Point (4, 1). For y-int, let x = 0 For y-int, let x = 0 5. x + 2y = 6 x y =3 (0, 3) 4. (0) + 2y = 6 (0) y = 3 3. 2. (6, 0). y=3 (0, 3) y = 3 (0, 3) 1. 0. (3, 0). -9 -8 -7 -6 -5 -4 -3 -2 -1. -1 0 1 2 3 4 5 6 7 8 9. For x-int, let y = 0 For x-int, let y = 0 (0, 3)-2. x + 2y = 6 x y=3 -3. x + 2 (0) = 6 x (0) = 3 -4. -5. x=6 (6, 0) x=3 (3, 0) -6. -7. -8. -9. Example 2: Find the solution of the following system of equations by using the graphing calculator. x + 2y = 6. x y =3. Step 1: Rearrange to Slope Step 2: Set WINDOWS to Standard Mode Step 3: Graph and y intercept form ZOOM GRAPH.

3 Y=. Choose Option 6. will set x: [ 10, 10, 1]. y: [ 10, 10, 1]. Step 4: Run INTERSECT Step 5: Verify using TABLE. 2nd CALC 2nd TABLE. ENTER ENTER ENTER GRAPH. TRACE. Choose Option 5. Page 2. Copyrighted by Gabriel Tang , PreCalculus Chapter 1: Linear and Quadratic Functions There are three types of solutions to a system of Linear equations: 1. Intersecting Lines 2. Parallel Lines 3. Overlapping Lines One distinct Solution No Solution Many (Infinite) Solutions Different Slopes Identical Slopes Identical Slopes Different y-Intercepts Different y-Intercepts Identical y-Intercepts 9 9 9. 8 8 8. 7 7 7. 6 6 6. 5 5 5. 4 4 4. 3 3 3. 2 2 2. 1 1 1. 0 0 0. -9 -8 -7 -6 -5 -4 -3 -2 -1. -1 0 1 2 3 4 5 6 7 8 9 -1 0 1 2 3 4 5 6 7 8 9. -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 0 1 2 3 4 5 6 7 8 9. -9 -8 -7 -6 -5 -4 -3 -2 -1. -2 -2 -2. -3 -3 -3. -4 -4 -4. -5 -5 -5. -6 -6 -6. -7 -7 -7. -8 -8 -8. -9 -9 -9. Example 3: Determine the number of solutions for the systems of equations below.

4 A. x + 2y = 10 b. 2x + 5y = 15. x + 2y = 6 6x +15y = 45. Line 1: Line 2: Line 1: Line 2: x + 2 y = 10 x + 2y = 6 2 x + 5 y = 15 6 x + 15 y = 45. 2 y = x + 10 2 y = x + 6 5 y = 2 x + 15 15 y = 6 x + 45. x + 10 x+6 2 x + 15 6 x + 45. y= y= y= y=. 2 2 5 15. 1 1 2 2. y= x+5 y= x+3 y= x+3 y= x+3. 2 2 5 5. 1 1 2 2. m= , y-int = 5 m = , y-int = 5 m= , y-int = 3 m= , y-int = 3. 2 2 5 5. Identical slopes and y-intercepts mean Identical slopes, but different y- overlapping lines. Therefore, this intercepts mean parallel lines. system has Therefore, this system has MANY SOLUTIONS. NO SOLUTION. When using the substitution method to solve a system of Linear equations: 1. Isolate a variable from one equation. (Always pick the variable with 1 as a coefficient.). 2. Substitute the resulting expression into that variable of the other equation. 3. Solve for the other variable. 4. Substitute the result from the last step into one of the original equation and solve for the remaining variable.

5 Copyrighted by Gabriel Tang , Page 3. Chapter 1: Linear and Quadratic Functions PreCalculus Example 4: Using the substitution method, algebraically solve the system of equations below. 2 (x 2y) = 5 (5 y). 3 (y x) = 2 (y + 7). Expand each equation accordingly. Solve for both variables using the substitution method. Equation 1 Isolate y from the first equation. 2 (x 2y) = 5 (5 y) 2x + y = 25. 2x 4y = 25 5y y = 25 2x 2x + y = 25 Substitute expression into y in the second equation. 5y 3x = 14. 5(25 2x) 3x = 14. Equation 2 125 10x 3x = 14. 3 (y x) = 2 (y + 7) 13x = 14 125 139. 13x = 139 x=. 3y 3x = 2y 14 13. Solve for the remaining variable. Pick the easier equation of the two. 5y 3x = 14 2x + y = 25. 139 . 2 + y = 25. 13 . 278. + y = 25. 13. 278 47. y = 25 y=. 13 13. Elimination by Multiplication: - most useful when neither equation has the same like terms. - by multiplying different numbers (factors of their LCM) on each equation, we can change these equations into their equivalent form with the same like terms.

6 3x + 7y = 3. Example 5: Solve the system of Linear equations 4y 5x = 42 by elimination. First, rearrange the equations so x and y terms will line up. 3x + 7y = 3. 5x + 4y = 42 Eliminate x by addition. Next, substitute y into one Eliminating x of the equations to solve 15x + 35y = 15 for x. LCM of 3x and 5x = 15x + ( 15x + 12y = 126) 3x + 7y = 3. 47y = 141 3x + 7 (3) = 3. (Multiply Equation 1 by 5 to obtain 15x). 141 3x = 3 21. (Multiply Equation 2 by 3 to obtain 15x) y=. 47 3x = 18. 5 (3x + 7y = 3) 15x + 35y = 15. y=3 x = 6. 3 ( 5x + 4y = 42) 15x + 12y = 126. Page 4. Copyrighted by Gabriel Tang , PreCalculus Chapter 1: Linear and Quadratic Functions Review of Coordinate Geometry x + x 2 y1 + y 2 . Distance d= ( x 2 x1 )2 + ( y 2 y1 )2 Midpoint M = 1 , . 2 2 . Example 6: Find the distance (in exact value) between A ( 3, 5) and B (7, 3). y 9. 8. d AB = (7 3)2 + ( 3 5)2. A ( 3, 5) 7. 6. = (10)2 + ( 8)2. (x1, y1). 5. = 100 + 64. 4. 3 = 164. 2. 1.

7 D AB = 2 41. 0 x -9 -8 -7 -6 -5 -4 -3 -2 -1-1 0 1 2 3 4 5 6 7 8 9. -2. -3. -4. -5. B (7, 3). -6 (x2, y2). -7. -8. -9. Example 7: A circle has a diameter with endpoints ( 4, 1) and (2, 6). Find the exact length of the radius. Let A (x1, y1) = ( 4, 1) and B (x2, y2) = (2, 6). diameter AB. Length of Diameter AB Length of the Radius =. 2. d AB = (2 4)2 + (6 1)2. 85. = 6 2 + 7 2 = 36 + 49 Length of the Radius =. 2. d AB = 85. Copyrighted by Gabriel Tang , Page 5. Chapter 1: Linear and Quadratic Functions PreCalculus Example 8: Find the midpoints of the following line segments CD where C (5, 4) and D ( 3, 8). y 9. 5 + 3 4 + 8 . 8 M CD = , . 7 2 2 . 6. C (5, 4) 2 4 . 5 = , . 4 2 2 . 3. 2 M CD = (1, 2). 1. 0 x -9 -8 -7 -6 -5 -4 -3 -2 -1-1 0 1 2 3 4 5 6 7 8 9. -2. M. -3. -4. -5. -6. -7. D ( 3, 8) -8. -9. Example 9: Given that the midpoint of AB is M (3, 3). If one of the endpoint of AB is A ( 3, 1), find the coordinate of endpoint B. 9 x + x2 y1 + y 2.

8 M AB = 2 , . 8 2 2 . 7. 6 (3, 3) = 3 + x2 , 1 + y 2 . 5 2 2 . 4. 3. A ( 3, 1) 2 3 + x2 1 + y2. 3= 3=. 1 2 2. 0 3 2 = 3 + x 2 3 2 = 1 + y2. -1. -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9. -2 M (3, 3) 6 = 3 + x 2 6 = 1 + y2. -3. 6 + 3 = x2 6 1 = y2. -4. -5 x2 = 9 y 2 = 7. -6 B. -7. -8. B = (9, 7). -9. 1-1 Assignment: pg. 5 7 #5 to 33 (every other odd) (Omit #9c e, 21). Page 6. Copyrighted by Gabriel Tang , PreCalculus Chapter 1: Linear and Quadratic Functions 1-2 and 1-3: Slopes of Lines and Finding Equations of Lines y9. 8 B (x2, y2) Slope: - a measure on the steepness of the line segment. 7. 6. 5 change in y (rise). 4. 3 y = (y2 y1) Slope of a Line Segment 2. 1. 0 rise change in y y -1. x m= = =. -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 run change in x x -2. -3. -4. y y -5. -6 m= 2 1. -7 change in x (run) x x 2 1. A (x1, y1) -8. -9 x = (x 2 x 1 ). Example 1: Find the slope of the following line segments. a. CD where C (5, 4) and D ( 3, 8) b.

9 PQ where P ( 6, 3) and Q (4, 2). 9 9. 8 8. 7 7. 6 6. 5 5. 4 C (5, 4) P ( 6, 3) 4. 3 3. 2 2. 1 1. 0 0. -1 0 1 2 3 4 5 6 7 8 9. -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 0 1 2 3 4 5 6 7 8 9. -9 -8 -7 -6 -5 -4 -3 -2 -1. -2 -2. -3 -3 Q (4, 2). -4 rise = (4 ( 8)) = 12 -4. -5 -5. -6 -6. -7 -7. -8 -8. D ( 3, 8). run =-9(5 ( 3)) = 8 -9. y 2 y1 y 2 y1. mCD = m PQ =. x 2 x1 x2 x1. 4 ( 8) 3 2 3 1. = m CD = = m PQ =. 5 ( 3) 2 4 ( 6 ) 2. 12 5. = =. 8 10. Copyrighted by Gabriel Tang , Page 7. Chapter 1: Linear and Quadratic Functions PreCalculus c. AB where A ( 5, 4) and B (3, 4) d. RS where R ( 3, 6) and S ( 3, 7). 9 9. 8 8. 7 7. 6. R ( 3, 6) 6. A ( 5, 4) 5 5. 4 B (3, 4) 4. 3 3. 2 2. 1 1. 0 0. -1 0 1 2 3 4 5 6 7 8 9. -9 -8 -7 -6 -5 -4 -3 -2 -1 -1 0 1 2 3 4 5 6 7 8 9. -9 -8 -7 -6 -5 -4 -3 -2 -1. -2 -2. -3 -3. -4 -4. -5 -5. -6 -6. -7 -7. -8 S ( 3, 7) -8. -9 -9. y 2 y1 y 2 y1. m AB = m RS =. x2 x1 x 2 x1. 4 4 7 6. = m AB = 0 = m RS = undefined 3 ( 5) 3 ( 3). 0 13. = =. 8 0.

10 In general, slopes can be classified as follows: y y x x Negative Slope (m < 0). Positive Slope (m > 0) Line goes DOWN from left to right. Line goes UP from left to right. y y x x Zero Slope (m = 0) Undefined Slope Horizontal (Flat) Line [Rise = 0] Vertical Line [Run = 0]. Page 8. Copyrighted by Gabriel Tang , PreCalculus Chapter 1: Linear and Quadratic Functions 3. Example 2: If the slope of a line is , and it passes through A (2t, t 2) and B (2, 3), find the value of t 4. and point A. y 2 y1. m= 4( t 1) = 3(2 2t ). x2 x1. 4t 4 = 6 6t 3 3 (t 2 ). = 4t + 6t = 6 + 4 t=5 A (2(5), (5) 2) = A (10, 3). 4 2 2t 2t = 10. 3 t 1. = 10. 4 2 2t t=. 2. 9 9. 8 8. 7 7. 6 6. 5 5. 4 4. 3 3. 2 2. 1 1. 0 0. -1 0 1 2 3 4 5 6 7 8 9. -9 -8 -7 -6 -5 -4 -3 -2 -1 -1. -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9. -2 -2. -3 -3. -4 -4. -5 -5. -6 -6. -7 -7. -8 -8. -9 -9. Parallel Lines Perpendicular Lines slope of line 1 = slope of line 2 slope of line 1 = negative reciprocal slope of line 2.