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sec1.3 Lines in plane,Slope Objective:: Review lines in a plane and slopes and business applications Slope intercept form of Equation of line y=mx+b m = slope and y intercept =0, b m= change_in_y y y −y = x = x 22 − x 11 = rise run change_in_x Vertical Line, slope undefined graph x = 2 Horizontal Line, slope = 0 graph y = 2 y (black) red 4 3 2 1 -4 -2 2 4 -1 x -2 Linear equation y = 2x + 1 Linear equation y = −x + 2 y positive slope (black) negative slope (red) 4 2 -2 -1 1 2 3 4 5 x -2 Linear equation x + y = 2 in slope intercept form solve for y, y = −x + 2 the same graph as the red on above. Example 2- Slope as ratio Maximum recommemded slope of wheelchair ramp is 1/12 = 0. 083 A business store ramp rises 22 in. overall (height) and a horizontal length of 24 in. Is the ramp steeper than recommended? ————————————— 1 vertical_change = 22in. = 0. 076 24in. horizontal_change 0.076 ≺ 0. 083 so the slope is not steeper than recommended. Solution: Slope m= Example 3 -Using Slope as a Rate of Change A manufacturing company determines the cost (C) in dollars of producing x units of C = 25x + 3500 Describe the practical significance of the y intercept and slope. ——————————————– Solution: The y intercept is (0,3500). The fixed costs of producing zero units is $3500, the cost paid regardless of the units produced Slope m=$25 is the cost of producing each unit. Economists call this the marginal costs If production increases by one unit,the "margin " or extra cost is $25 C = 25x + 3500 y 8000 7000 6000 5000 4000 3000 2000 1000 0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 x Production Costs ———————————— Objective: Examine a Business application of Break-Even analysis Include Sec 1.2 # 61 as part of your homework Possible Test Question Sec 1.2 Example 5 Finding a Break-Even Point (p49) A business manufactures a product at a cost of $0.65 per unit and sells the product for $1.20 per unit. The company’s initial investment was $10,000 to produce the product (fixed costs) a.Will the company break even if it sells 18,000 units? b. How many units must the company sell to break even? ————————————————2 C = total cost of producing x units R= revenue from the sale of x units p = selling price per unit C=mx+b where m = cost per unit of each item x and b = initial investment and fixed costs R=px where p = selling price per unit The break even point is where Revenue=Costs R=C ——————————– a. Write the cost and revenue equations C = 0. 65x + 10, 000 R = 1. 20x Write the break even equation R=C 1. 20x = 0. 65x + 10, 000 Substitute x=18,000 into the equation to see if R=C R C 1. 20 ⋅ 18000 ? 0. 65 ⋅ 18000 + 10000 21600 21700 Costs are greater than the Revenue so the company will not break even if it sells 18,000 units b. What is the break even point? Solve the break even equation for x R=C 1. 20x = 0. 65x + 10, 000 Solve for x, the number of units 1. 20x − 0. 65x = 10000, x = 18, 182 the number of units sold before there is any profit To find the revenue or costs at this value evaluate R = px R = 1. 20x = 1. 20 ⋅ 18182 = $21818. Break even point where R = C is 18182, 21818 C= (black) R= (red) 3 y 40000 30000 20000 10000 0 0 10000 20000 30000 x Loss before the break even point is C−R costs are greater than sales Profit after the break even point is P = R−C sales exceed costs where P = profit ———————————– Back to Sec 1.3 Finding the Slope of a Line y change_in_y y −y = x = x 22 − x 11 = rise slope m = run change_in_x where point 1 =x 1 , y 1 point 2 =x 2 , y 2 Exampe 4-Finding the slope of a Line Find the slope of the line passing through each pair of points a. (-2,0) and (3,1) y −y 1 − 0 m= x 22 − x 11 = = 1 = 1 5 3+2 3 − −2 b. (-1,2) and (2,2) y −y m= x 22 − x 11 = 2 − 2 = 0 = 0 3+1 3 − −1 y 4 2 -2 2 4 x -2 y = 2_horizontal_line c. (0,4) and (1,-1) y −y m= x 22 − x 11 = −1 − 4 = −5 = −5 1−0 1 d. (3,4) and (3,1) y −y m= x 22 − x 11 = 1 − 4 = −3 = undefined 3−3 0 4 because division by zero is undefined 3, −3, 3, 5 y 4 2 -2 2 4 -2 x x = 3_vertical_line Point Slope form of the line y − y 1 = mx − x 1 Example 5-Using Point Slope form of line Find the equation of the line that has slope 3 and passes thru the point (1,-2) Answer in slope intercept form and standard form. Solution: y − y 1 = mx − x 1 y − −2 = 3x − 1 y + 2 = 3x − 3 y = 3x − 5 slope_intercept_form standard form ax + by = c standard_form 3x − y = 5 Parallel (∥)and Perpendicular⊥ lines Similar to Example 7 in text Parallel Lines Two lines are parallel (∥)if and only if they have the same slope or are both vertical. Example Find an equation of the line through the point −3, −2 parallel to the line 2x − 7y + 6 = 0 Solution a. All you need to know is the slope of the line 2x − 7y + 6 = 0. Solving for y gives −7y = −2x − 6. y = 27 x + 67 m = 27 (using the slope-intercept form of the line). b. Find the equation (in point-slope form) m= 27 Use y − y 1 = mx − x 1 y − −2 = 27 x − −3 y + 2 = 27 x + 3 Multiplying both sides of the equation by 7 and using the distributive property gives 5 7y + 14 = 2x + 6 standard form 0 = 2x − 7y − 8 or 2x − 7y − 8 = 0 Note: Most texts use a positive x term first Perpendicular Lines Two lines with nonzero slopes m 1 and m 2 are perpendicular⊥ if and only if m 1 m 2 = −1, that is, if their slopes are negative reciprocals (m 2 = − m11 ). Also, any horizontal line (slope 0) is perpendicular to any vertical line (slope undefined). Example Find an equation of the line through the point −3, −2 perpendicular to the line 2x − 7y + 6 = 0 Solution a. Find the slope of the given line: Example 2x − 7y + 6 = 0 gives ,as above, m= So the slope of the given line is 2 7 2 7 . The slope of a line perpendicular to this is then the negative reciprocal m 2 = − m11 − 12 = − 72 7 b. You want the equation of a line through −3, −2 with a slope of − 72 : Use y − y 1 = mx − x 1 y − −2 = − 72 x − −3 y + 2 = − 72 x + 3 Multiplying both sides of the equation by −2 and using the distributive property gives − 2y − 4 = 7x + 21 0 = 7x + 2y + 25 or 7x + 2y + 25 = 0 2 6 original y = 7 x + 7 blue parallel 2x − 7y − 8 = 0black perpendicular 2x − 7y + 6 = 0. y = − 72 x − 252 red 6 y 4 2 -6 -5 -4 -3 -2 -1 1 2 3 4 5 x -2 Example 8 - Depreciating equipment A ocmpany purchased a $12,000 machine with a useful life of 8 yrs. The salvage value is then $2000. a. Write a linear equation that describes the book value of the machine each year. b. Show a table for each year using this equation .——————————– Solution: Use V = mt + b where V = value of the machine at any time t m = depreciation rate in dollars per year b= value at t = 0 t 0 8 V 12,000 2000 m= V 2 − V 1 = 2000 − 12000 = −1250 t2 − t1 8−0 depreciation rate in dollars/yr. noted by the negative slope Using slope intercept equation V=mt+b V = −1250t + 12000 y 12000 10000 8000 6000 4000 2000 0 0 2 4 6 8 x b. V = −1250t + 12000 7 t 0 1 2 3 4 5 6 7 8 V 12,000 10750 9500 8250 7000 5750 4500 3250 2000 8