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MAT091 :: Lecture Note :: Week 10
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(due Tuesday, 10/31/2017)
(due Tuesday, 10/31/2017)
MW.com defines linear as follows.
"of, relating to, resembling, or having a graph that is a line and especially a straight line" [...and...] "having or being a response or output that is directly proportional to the input"Observe that the word linear contains the word line.
A linear function is a function is a "first degree polynomial function of one variable. These functions are called 'linear' because they are precisely the functions whose graph in the Cartesian coordinate plane is a straight line."
The following are equations for a line.
slopeintercept form: y = mx + b ... used when slope and yintercept are known pointslope form: y  y_{1} = m(x  x_{1}) ... where (x_{1}, y_{1}) is a point on the line having slope m standard form: Ax + By = CSlope is another way of saying "rate of change" and linear functions have a constant rate of change.
Slope is often described as the ratio "rise over run."
Linear functions with a positive constant rate of change are increasing functions. Linear functions with a negative constant rate of change are decreasing functions.
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The slope of a line is a measure of its average rate of change (steepness?). The slope also indicates if the line is increasing (uphill?) or decreasing (downhill?). In linear equations, slope is represented by the letter m.
The following is how to find the slope between two unique points on a line.
point 1: (x_{1}, y_{1}) point 2: (x_{2}, y_{2}) slope = m = (y_{2}  y_{1}) / (x_{2}  x_{1}) note: x_{1} ≠ x_{2}Slope represents the ratio of changeinoutput over changeininput.
m = Δy / Δx (change in output / change in input) Δ is the "delta" character Δy is read "change of" (Δy is read "change of" y) Δy = y_{2}  y_{1}Slope is often described as the ratio "rise over run."
rise m =  runSome slope notes.
horizonal lines have slope 0 (y = f(x) = k, where k is a consant) vertical lines have undefined slope (x = k, where k is a consant) parallel lines have equal slopes (m_{1} = m_{2}) perpendicular lines have negative inverse slopes (m_{2} = 1/m_{1})The following is a verbose form the slopeintercept equation.
output = slope * input + initial_value_if_any ...or... output = average_rate_of_change * input + initial_value_if_any
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We frequently need to know the points where lines (graphs in general) intersect (cross) the horizonalaxis and verticalaxis (often called the xintercept and yintercept, respectively).
The verticalintercept for a function is found by evaluating the function when its input is zero. The orderdpair for a verticalintercept will be
(0, something)
.The horizontalintercept for a function is found by setting its output to zero and finding what input value results in that output. The orderdpair for a horizonalintercept will be
(something, 0)
.f(x) = 5(x) + 3 verticalintercept (yintercept)... set the input to 0 and evaluate: f(0) = 5(0) + 3 = 3 verticalintercept is (0, 3) horizontalintercept (xintercept)... set the output to 0 and solve: 0 = 5(x) + 3 3 = 5(x) 3/5 = x horizontalintercept is (3/5, 0)Generalizations.
given f(x) = m(x) + b verticalintercept: (0, b) [notice input is 0] horizontalintercept: (b/m, 0) [notice output is 0]
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Parallel lines never intersect (i.e. they don't have any points in common) and they have equal slopes.
Given the line
y = 10x + 2
find the equation of a parallel line that passes through the point(5, 40)
.y = 10x + b # parallel lines have the same slope y  40 = 10(x  5) # using pointslope form y  40 = 10x  50 # apply distributive property y  40 + 40 = 10x  50 + 40 # add 40 to both sides y = 10x  10 # y = mx + b form [check answer] substitute 5 in for x and we should get 40 y = 10(5)  10 = 50  10 = 40
[question]
No or Yes: The parallel line intercepts the verticalaxis at the point(0, 10)
.
[question]
No or Yes: The parallel line intercepts the horizontalaxis at the point(1, 0)
.
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Perpendicular lines intersect (i.e. they have a point in common) and they form right angles at their point of intersection.
The slopes of perpendicular lines are negative reciprocals of each other.
m * m_{p} = 1 'm' is the slope of a line and 'm_{p}' is the slope of a perpendicular lineGiven the line
y = 5x + 2
, find the equation of a perpendicular line that passes through the point(5, 10)
.Let A be the given line and B the perpendicular line. slope of line A: m = 5 slope of line B: m = (1/5) # negative reciprocal find the slopeintercept equation for line B substitute point (5, 10) into pointslope equation... y  10 = (1/5)(x  5) solve for y... y  10 + 10 = (1/5)(x  5) + 10 # add 10 to both sides y = (1/5)(x  5) + 10 # simplify y = (1/5)x  (1/5)5 + 10 # distribute (1/5) y = (1/5)x + 1 + 10 # simplify y = (1/5)x + 11
[question]
No or Yes: The perpendicular line intercepts the verticalaxis at the point(0, 11)
.
[question]
No or Yes: The perpendicular line intercepts the horizontalaxis at the point(55, 0)
.
[exercise]
Find the point where lines A and B intersect.set line A equal to line B 5x + 2 = (1/5)x + 11 solve for x... 5x + 2 = 0.2x + 11 # divide 1 by 5 5x + 2  2 = 0.2x + 11  2 # subtract 2 from both sides 5x = 0.2x + 9 # simplify 5x + 0.2x = 9 # add 0.2x to both sides 5.2x = 9 # simplify 5.2x / 5.2 = 9 / 5.2 # divide both sides by 5.2 x = 1.73 # simplify substitute x in one of the equations... y = 5(1.73) + 2 = 10.65 point of intersection: (1.73, 10.65) check using the other line... (1/5)1.73 + 11 = 10.65
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Two lines that are not parallel can intersect. If the equations for the lines are known, then the point of intersection can be determined as follows.
line A: y = 2x + 2 line B: y = 5x + 5 set the lines equal to one another... 2x + 2 = 5x + 5 solve for x... 2x + 2 + 2 = 5x + 5 + 2 # add 2 to both sides 2x = 5x + 3 2x + 5x = 5x + 3 + 5x # add 5x to both sides 3x = 3 3x / 3 = 3 / 3 # divide both sides by 3 x = 1 substitute for x in one of the line equations and evaluate... y = 2(1) + 2 = 2 + 2 = 0 point of intersection: (1, 0) check answer: plug (1, 0) into the other line equation and see if it works... y = 5(1) + 5 = 5 + 5 = 0
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Zelmo Zeroman had to select between two linear pay scales. One choice was to receive $10 upfront and then earn $1 for every 8 hours worked. The second choice was to received $2 upfront and then earn $1 for every 5 hours worked. The following two functions define Zelmo's pay scale choices.
1 1 p(h) = (h) + 10 P(h) = (h) + 2 8 5 h: number of hours p(h): USD P(h): USD The amount of pay is a function of hours worked. The amount of pay depends on the number of hours worked.Zelmo realized that over time the
P(h)
pay scale would result in more pay, but he wanted to learn how many hours he needed to work beforeP(h)
would be greater thanp(h)
.It was suggested that Zelmo find where
P(h)
andp(h)
intersect when graphed.observe... 1/8 = 0.125 and 1/5 = 0.2 p(h) = 0.125(h) + 10 and P(h) = 0.2(h) + 2 let p(h) = P(h)... 0.125(h) + 10 = 0.2(h) + 2 solve for 'h'... ...subtract 0.125(h) from both sides... 0.125(h) + 0.125(h) + 10 = 0.125(h) + 0.2(h) + 2 10 = 0.075(h) + 2 ...subtract 2 from both sides... 2 + 10 = 0.075(h) + 2  2 8 = 0.075(h) ...divide both sides by 0.075... 8/0.075 = 0.075(h)/0.075 106.67 = hConclusion: Zelmo should pick the
p(h)
pay scale if he is going to work less than106.67 hours
; otherwise, he should select the pay scale defined byP(h)
.
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Zelmo Zeroman borrowed $1,000 from his "friend" Herb Mumford. Zelmo and Herb agreed that the loan was to be paid off at a rate of $200 per week. Zelmo created the following function to help him keep track of his weekly payoff amounts.
p(w) = 1000  200(w) USD input 'w' is week number and USD stands for U.S. Dollar domain: w ≥ 0, where 'w' is an integer range: 0 ≤ p(w) ≤ 1000The function p(w) outputs the payoff amount after w weeks.
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The following are some inequalities graphed on the number line. An endpoint marked with 'x' (or ']') implies that value is included, while an endpoint marked with 'o' (or ')') is excluded.
x > 2 o=========> 5 4 3 2 1 0 1 2 3 4 5 x ≥ 4 x===========================> 5 4 3 2 1 0 1 2 3 4 5 x ≤ 3 <=========================x 5 4 3 2 1 0 1 2 3 4 5 x < 2 <==========o 5 4 3 2 1 0 1 2 3 4 5Linear inequalities are solved just like linear equations with one exception: the inequality sign is "flipped" whenever there is a multiply (or divide) by a negative. The term "flipped" is used to imply lessthan becomes greaterthan and vice versa; and that lessthanorequalto becomes greaterthanorequalto and vice versa.
Example 1: 8x + 1 < 2x  5 8x  2x + 1 < 5 6x + 1  1 < 5  1 6x < 6 x < 1 try x = 0... 8(0) + 1 = 9; 2(0)  5 = 5 9 is not less than 5 try x = 1... 8(1) + 1 = 7; 2(1)  5 = 7 7 is not less than 7 try x = 2... 8(2) + 1 = 15; 2(2)  5 = 9 15 is less than 9 Example 2: 2x < 5 2x / 2 > 5 / 2 # < flipped to > (divideby negative) x > 2.5 try x = 5... 2(5) < 5 10 < 5 is true try x = 5... 2(5) < 5 10 < 10 is falseThe following uses numbers to help understand when inequalities are flipped (switched).
10 > 5 is true multiply both sides by 1 10(1) = 10 and 5(1) = 5 10 > 5 is falseExternal Hyperlink(s)
 PurpleMath.com:: Graphing Linear Inequalities: y > mx + b, etc.
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A "system" of equations is a collection of equations that are treated as an atomic unit. In other words, a system of equations is used when dealing with more than one variable.
A: 3x + 3y = 0 B: 12x + 2y = 20 pick one of the equations... pick A solve equation A for y 3x + 3y = 0 3x + 3x + 3y = 0 + 3x 3y = 3x 3y / 3 = 3x / 3 y = x substitute for y in the equation B... 12x + 2(x) = 20 solve equation B for x... 12x + 2x = 20 10x = 20 10x / 10 = 20 / 10 x = 2 pick one of the equations... pick A substitute for x in the equation A... 3(2) + 3y = 0 solve equation A for y... 3(2) + 3y = 0 6 + 3y = 0 6 + 6 + 3y = 0 + 6 3y = 6 3y / 3 = 6 / 3 y = 2 the solution is the point (2, 2)Another system of linear equations.
Two numbers have a sum of 55 and a difference of 9. Find the numbers. let x and y be the two unknown numbers form two equations... A: x + y = 55 B: x  y = 9 add the two equations together... x + y = 55 + x  y = 9 ============= 2x = 64 solve for x... 2x = 64 2x / 2 = 64 / 2 x = 32 pick an equation, substitute for x, solve for y pick A 32 + y = 55 32 + 32 + y = 55 + 32 y = 23 solution: (32, 23) [exercise] Redo this problem using substitution.Here is another system of linear equations that is solved using the addition method.
A: 5x + y = 10 B: x + y = 20 multiply both sides of B by 1... 1(x + y) = 20(1) x  y = 20 add the two equations... 5x + y = 10 + 1x  y = 20 =============== 4x = 10 solve for x... 4x / 4 = 10 / 4 x = 2.5 pick an equation and solve for y... pick B 2.5 + y = 20 y = 22.5 solution: (2.5, 22.5) subtitute the solution into A and see if it works... 5(2.5) + 22.5 = 10 12.5 + 22.5 = 10 it works!Is a system of equations necessary to solve the following problem?
Ten less than half of a number is thirty. What is the number?No, because there is only one variable.
n/2  10 = 30 n/2  10 + 10 = 30 + 10 n/2 = 40 n/2 * 2 = 40 * 2 n = 80Here is a exercise that can be solved using a system of equations.
A new nightclub had a grand opening in Scottsdale. First night admission was $4 for females and $6 for males. 90 people attended the grand opening and the nightclub collected $460. How many females and how many males attended the grand opening?let 'f' represent # of females and let 'm' represent # of males A: f + m = 90 B: $4(f) + $6(m) = $460 pick A and solve for 'f'... f = 90  m substitute 'f' into B... 4(90  m) + 6m = 460 distribute the 4, combine liketerms, solve for 'm'... 360  4m + 6m = 460 360 + 2m = 460 360 + 360 + 2m = 460  360 2m = 100 (1/2)2m = 100(1/2) m = 50 substitute m = 50 into A and solve for 'f'... f + 50 = 90 f = 40 There were 40 females and 50 males the grand opening.Education.Yahoo.com:: Solving Linear Systems
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The "intersection" feature of the TI83 calculator can be used to find the solution to a system of linear equations.
press Y= Y_{1} press CLEAR, if necessary enter equation: X is entered using key labeled X,T,theta,n 2X + 3 ENTER cursor at Y_{2}, press CLEAR if necessary enter equation: (1/3)X  4 ENTER press GRAPH press 2ND TRACE 5 (intersect) First curve? ENTER Second curve? ENTER Guess? ENTER Intersection x = 3 and y = 3 (3, 3)Substitute
x = 3
into both equations and test if(3, 3)
is a solution.f(x) = 2x + 3 f(3) = 2(3) + 3 = 6 + 3 = 3 check f(x) = (1/3)x  4 f(3) = (1/3)(3)  4 = 1  4 = 3 check
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