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Evaluate each expression for x = –1, 0, and 2 .

2. – 1. 2. x. 3. 5. 1. –. , –1,. ANSWER. 3. 3. Warm Up #8. Evaluate each expression for x = –1, 0, and 2. 1. 2 x + 3. 1, 3, 7. ANSWER. Graphing Lines. Method 1: From the slope-intercept form y = mx + b. Ex. Graph y = 2x - 3. 1. y axis.

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Evaluate each expression for x = –1, 0, and 2 .

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  1. 2 – 1 2. x 3 5 1 – , –1, ANSWER 3 3 Warm Up #8 Evaluate each expression forx=–1, 0,and2. 1. 2x + 3 1, 3, 7 ANSWER

  2. Graphing Lines Method 1: From the slope-intercept form y = mx + b Ex. Graph y = 2x - 3 1 y axis Start on the y axis Go from that point up or down (depending if it’s – or +) the numerator, then over to the right the denominator x axis In this case, up two, over one Draw a line through the two points

  3. Graph both lines on the same graph: y = -2x + 4 and y = 3x - 2

  4. Graph both lines on the same graph: y = -¾x – 2 and y = -¾x + 2 Same slope: Parallel Lines

  5. Graph both lines on the same graph: y = ½x – 2 and y = -2x + 2 Negative Reciprocal slopes: Perpendicular Lines Lines

  6. 2. y = x – 2 for Examples 1 and 2 GUIDED PRACTICE Graph the equation. Compare the graph with the graph of y = x. SOLUTION y = x The graphs ofy = x – 2 and y = x both have a slope of 1, but the graph ofy = x – 2has a y-intercept of –2instead of 0. y = x - 2

  7. 3. y = 4x for Examples 1 and 2 GUIDED PRACTICE Graph the equation. Compare the graph with the graph of y = x. SOLUTION y = x The graphs ofy = 4x andy = x both have a y-intercept of 0, but the graph ofy = 4x has a slope of4 instead of 1. y = 4x

  8. 2 5 5. y = x + 4 for Examples 1 and 2 GUIDED PRACTICE Graph the equation 4. y = –x + 2

  9. for Examples 1 and 2 GUIDED PRACTICE Graph the equation 9. f (x) = -2 + x 8. f (x) = 1 – 3x f(x) = x – 2 f(x) = -3x + 1

  10. EXAMPLE 3 Solve a multi-step problem Biology The body length y (in inches) of a walrus calf can be modeled by y = 6x + 48 where xis the calf’s age (in months). • Graph the equation. • Describe what the slope and y-intercept represent in this situation. • Use the graph to estimate the body length of a calf that is 10 months old.

  11. EXAMPLE 3 Solve a multi-step problem SOLUTION STEP 1 Graph the equation. (2 , 60) y = 6x + 48 (0 , 48) STEP 2 Interpret the slope and y-intercept. The slope, 6, represents the calf’s rate of growth in inches per month. The y - intercept, 48, represents a newborn calf’s body length in inches.

  12. EXAMPLE 3 Solve a multi-step problem STEP 3 Estimate the body length of the calf at age 10 months by starting at 10 on the x-axis and moving up until you reach the graph. Then move left to the y-axis.At age 10 months, the body length of the calf is about 115 inches.

  13. Method 2: Using two intercepts from Standard Form ax + by = c Graph: 3x + 4y = -12 x intercept ( , 0 ) let y = 0 and solve for x -4 3x + 4(0) = -12 3x = -12 x = -12 = - 4 3 y intercept ( 0 , ) let x = 0 and solve for y -3 3(0) + 4y = -12 4y = -12 y = -12 = - 3 4

  14. After finding the two intercepts, plot both points on the graph and draw a line through them x intercept ( -4 , 0 ) y intercept ( 0 , -3 )

  15. Graph both lines on the same graph: 3x – 2y = 6 and 5x + 2y = 10 Line 1: x int. ( , 0 ) 2 y int. ( 0 , ) -3 Line 2: x int. ( , 0 ) 2 y int. ( 0 , ) 5

  16. EXAMPLE 5 Graph horizontal and vertical lines Graph (a) y = 2 and (b) x = –3. SOLUTION a. The graph of y = 2 is the horizontal line that passes through the point (0, 2). Notice that every point on the line has a y-coordinate of 2. b. The graph of x = –3 is the vertical line that passes through the point (–3, 0). Notice that every point on the line has an x-coordinate of –3.

  17. for Examples 4 and 5 GUIDED PRACTICE Graph the equation. 14. y = –4 13. x = 1

  18. 3 4 3 y =x +(–2) Substitute for mand –2 for b. 4 3 4 EXAMPLE 1 Write an equation given the slope and y-intercept SOLUTION Given the y-intercept is -2 and the slope is ¾ Use the slope-intercept form to write an equation of the line y =mx +b Use slope-intercept form. y =x – 2 Simplify.

  19. 7 7 3 3. m = – , b = 2 2 4 3 y =– x + 4 y =x + 1 3 for Example 1 GUIDED PRACTICE Write an equation of the line that has the given slope and y-intercept. 1. m = 3, b = 1 ANSWER ANSWER 2. m = –2 , b = –4 ANSWER y = –2x – 4

  20. To write the equation of a line in slope-intercept form y = mx + b x1 y1 Point ( , ) Slope m = • Identify a point and a slope • Substitute values into point-slope form • Solve for y y – y1 = m(x – x1) y = mx + b Answer will be in slope-intercept form Write an equation given the slope and a point

  21. x1 y1 Point ( , ) Slope m = 4 -2 Write the equation of the line with the given information shown 3/4 y – y1 = m(x – x1) • Identify a point and a slope • Substitute values into point-slope form • Solve for y y – (-2) = 3/4(x – 4) y + 2 = 3/4x – 3 y = 3/4x – 5

  22. EXAMPLE 2 Write an equation given the slope and a point Write an equation of the line that passes through (5, 4) and has a slope of –3. x1 y1 Point ( , ) Slope m = 5 4 -3 y –y1=m(x –x1) y –4=–3(x –5) y – 4 = –3x + 15 y = –3x + 19

  23. Write equations of parallel or perpendicular lines EXAMPLE 3 Write an equation of the line that passes through (–2,3) and is (a) parallel to the line y= –4x + 1. x1 y1 Point ( , ) Slope m = -2 3 Remember: Parallel lines have the same slope -4 y –y1=m(x –x1) y –3=–4(x –(-2)) y –3=–4(x + 2)) y –3=–4x– 8 y = –4x – 5

  24. Write equations of parallel or perpendicular lines EXAMPLE 3 Write an equation of the line that passes through (–2,3) and is (b) perpendicular to, the line y= –4x + 1. x1 y1 Point ( , ) Slope m = -2 3 Remember: Perpendicular lines have the negative reciprocal slopes ¼ y –y1=m(x –x1) y –3=¼(x –(-2)) y –3=¼(x + 2)) y –3=¼x + ½ y = ¼x + 3 ½

  25. 12 –3 10 – (–2) y2 – y1 = m= = = –4 x2 – x1 2 –5 EXAMPLE 4 Write an equation given two points Write an equation of the line that passes through (5, –2) and (2, 10). SOLUTION The line passes through (x1, y1) = (5,–2) and (x2, y2) = (2, 10). Find its slope.

  26. EXAMPLE 4 Write an equation given two points You know the slope and a point on the line, so use point-slope form with either given point to write an equation of the line. x1 y1 Point ( , ) Slope m = 2 10 -4 y– y1=m(x –x1) y–10=–4(x –2) y –10 = – 4x + 8 y = – 4x + 18

  27. Classwork Workbook 2-3 (1-30 multiples of 3) Workbook 2-4 (1-23 odd)

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