Day 8 Lines and Graphs Study Notes

Lines and Graphs

  • Key terms: y-intercept, x-intercept, origin (0,0).

  • Visual cue: a line with positive slope rises from left to right.

  • A line is determined by its slope and intercepts; understanding intercepts helps with graphing.

  • Slope concept:

    • Slope measures steepness of a line.
    • Positive slope: line goes up as you move left to right.
    • Negative slope: line goes down as you move left to right.
    • Slope formula (rise over run):
    • m=riserun=y<em>2−y</em>1x<em>2−x</em>1m = \frac{\text{rise}}{\text{run}} = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1}

  • Example: find the slope of the line with points (1,1) and (3,5).

    • m=5−13−1=42=2m = \frac{5 - 1}{3 - 1} = \frac{4}{2} = 2
    • The order of the points does not matter as long as you are consistent: swapping points yields the same slope value.

Slope

  • Slope is a measure of steepness; it is the rate of change of y with respect to x.
  • Calculation reminder:
  • m=y<em>2−y</em>1x<em>2−x</em>1m = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1}
  • For the pair (1,1) and (3,5), the slope is m=2].</li></ul><h3id="horizontalandverticallines">HorizontalandVerticalLines</h3><ul><li>Horizontallines:<ul><li>Equation:m = 2].</li> </ul> <h3 id="horizontalandverticallines">Horizontal and Vertical Lines</h3> <ul> <li>Horizontal lines:<ul> <li>Equation:y = kwherekisaconstant.</li><li>Slope:where k is a constant.</li> <li>Slope:m = 0.</li><li>Allpointsonthelinehavethesamey−value(k).</li></ul></li><li>Verticallines:<ul><li>Equation:.</li> <li>All points on the line have the same y-value (k).</li></ul></li> <li>Vertical lines:<ul> <li>Equation:x = kwherekisaconstant.</li><li>Slopeisundefined(infinite).Thiscorrespondstoaverticalrisewithzerorun.</li><li>Allpointsonthelinehavethesamex−value(k).</li></ul></li></ul><h3id="equationsoflines">EquationsofLines</h3><ul><li>Generalformofaline:where k is a constant.</li> <li>Slope is undefined (infinite). This corresponds to a vertical rise with zero run.</li> <li>All points on the line have the same x-value (k).</li></ul></li> </ul> <h3 id="equationsoflines">Equations of Lines</h3> <ul> <li>General form of a line:y = mx + b<ul><li>m=slope,b=y−intercept(thevalueofywherethelinecrossesthey−axis).</li><li>Slope−interceptformshowsthedirectrelationshipbetweenxandywiththeinterceptonthey−axis.</li></ul></li><li>Example:Graphthelinegivenby<ul> <li>m = slope, b = y-intercept (the value of y where the line crosses the y-axis).</li> <li>Slope-intercept form shows the direct relationship between x and y with the intercept on the y-axis.</li></ul></li> <li>Example: Graph the line given byy = -x + 2.<ul><li>Slope:<ul> <li>Slope:m = -1,intercept:, intercept:b = 2.</li></ul></li></ul><h3id="xinterceptandyintercept">X−InterceptandY−Intercept</h3><ul><li>X−intercept:thepointwherethelinecrossesthex−axis;set.</li></ul></li> </ul> <h3 id="xinterceptandyintercept">X-Intercept and Y-Intercept</h3> <ul> <li>X-intercept: the point where the line crosses the x-axis; sety = 0andsolveforx.<ul><li>Forthelineand solve for x.<ul> <li>For the liney = -x + 2,solve, solve0 = -x + 2togetto getx = 2;x−interceptis(2,0).</li></ul></li><li>Examplewithtwopoints:linethrough(0,−3)and(4,0).<ul><li>Slope:; x-intercept is (2, 0).</li></ul></li> <li>Example with two points: line through (0, -3) and (4, 0).<ul> <li>Slope:m = \frac{0 - (-3)}{4 - 0} = \frac{3}{4}.</li><li>Equationinslope−interceptform:using(0,−3)gives</li> <li>Equation in slope-intercept form: using (0, -3) givesy = \frac{3}{4}x - 3.</li><li>X−intercept:sety=0:</li> <li>X-intercept: set y = 0:0 = \frac{3}{4}x - 3 \Rightarrow x = 4.</li><li>Y−intercept:atx=0,y=−3,soy−interceptis−3.</li></ul></li></ul><h3id="pointslopeform">Point−SlopeForm</h3><ul><li>Ifalinehasslope</li> <li>Y-intercept: at x = 0, y = -3, so y-intercept is -3.</li></ul></li> </ul> <h3 id="pointslopeform">Point-Slope Form</h3> <ul> <li>If a line has slopemandpassesthroughapoint((x<em>1,y</em>1)),itsequationis:</li><li>and passes through a point ((x<em>1, y</em>1)), its equation is:</li> <li>y - y1 = m(x - x1)</li><li>Example:Findtheequationofthelinethrough((−1,1))and((3,7)).<ul><li>Slope:</li> <li>Example: Find the equation of the line through ((-1, 1)) and ((3, 7)).<ul> <li>Slope:m = \frac{7 - 1}{3 - (-1)} = \frac{6}{4} = \frac{3}{2}.</li><li>Usingpoint((−1,1)):</li><li></li> <li>Using point ((-1, 1)):</li> <li>y - 1 = \frac{3}{2}(x - (-1)) = \frac{3}{2}(x + 1).</li><li>Toslope−interceptform,simplify:</li><li></li> <li>To slope-intercept form, simplify:</li> <li>y = \frac{3}{2}x + \frac{5}{2}.</li></ul></li></ul><h3id="convertingtoslopeinterceptform">ConvertingtoSlope−InterceptForm</h3><ul><li>Frompoint−slope:</li></ul></li> </ul> <h3 id="convertingtoslopeinterceptform">Converting to Slope-Intercept Form</h3> <ul> <li>From point-slope:y - y1 = m(x - x1),solvefory:</li><li>, solve for y:</li> <li>y = mx + (y1 - m x1).</li><li>Examplerecap:with(m=32),(x<em>1=−1),(y</em>1=1):</li><li>Intercept</li> <li>Example recap: with (m = \frac{3}{2}), (x<em>1 = -1), (y</em>1 = 1):</li> <li>Interceptb = y1 - m x1 = 1 - \frac{3}{2}(-1) = 1 + \frac{3}{2} = \frac{5}{2}.</li><li>Therefore,</li> <li>Therefore,y = \frac{3}{2}x + \frac{5}{2}.</li></ul><h3id="standardform">StandardForm</h3><ul><li>Standardform:</li> </ul> <h3 id="standardform">Standard Form</h3> <ul> <li>Standard form:A x + B y = CwhereA,B,Careintegers(oftenwithgcd(A,B,C)=1).</li><li>Tofindslopefromstandardform,solveforytogetslope−interceptform:</li><li>where A, B, C are integers (often with gcd(A,B,C) = 1).</li> <li>To find slope from standard form, solve for y to get slope-intercept form:</li> <li>A x + B y = C \Rightarrow B y = -A x + C \Rightarrow y = -\frac{A}{B}x + \frac{C}{B}.</li><li>Theslopeis</li> <li>The slope ism = -\frac{A}{B}.</li><li>Example:2x+3y=1.<ul><li>Solvefory:</li> <li>Example: 2x + 3y = 1.<ul> <li>Solve for y:3y = -2x + 1 \Rightarrow y = -\frac{2}{3}x + \frac{1}{3}.</li><li>Slope:</li> <li>Slope:m = -\frac{2}{3}.

Applied Example: Linear Depreciation Model

  • Problem: Mark bought a car for $20,000 and it will have a trade-in value of $6,000 in 10 years. Assuming a constant rate of depreciation (linear), find a linear model describing the value after t years.
  • Given:
    • When t = 0,,V = 20{,}000.</li><li>When</li> <li>Whent = 10,,V = 6{,}000.</li></ul></li><li>Slope(rateofdepreciation):<ul><li></li></ul></li> <li>Slope (rate of depreciation):<ul> <li>m = \frac{6{,}000 - 20{,}000}{10 - 0} = \frac{-14{,}000}{10} = -1{,}400.</li></ul></li><li>Intercept(valueatt=0):</li></ul></li> <li>Intercept (value at t = 0):b = 20{,}000.</li><li>Linearmodel(valueasafunctionoftime):<ul><li></li> <li>Linear model (value as a function of time):<ul> <li>V(t) = -1{,}400\, t + 20{,}000.</li></ul></li><li>Verification:<ul><li></li></ul></li> <li>Verification:<ul> <li>V(0) = -1{,}400\cdot 0 + 20{,}000 = 20{,}000.</li><li></li> <li>V(10) = -1{,}400\cdot 10 + 20{,}000 = -14{,}000 + 20{,}000 = 6{,}000.</li></ul></li></ul><h3id="keytakeawaysandconnections">KeyTakeawaysandConnections</h3><ul><li>Theslopemisthecornerstoneoflinearrelationships;itdictatesdirectionandsteepnessofthegraph.</li><li>Intercepts,bothx−andy−,anchorthelineontheaxesandaidinquickgraphing.</li><li>Multipleformsexistforrepresentinglines:<ul><li>Slope−intercept:</li></ul></li> </ul> <h3 id="keytakeawaysandconnections">Key Takeaways and Connections</h3> <ul> <li>The slope m is the cornerstone of linear relationships; it dictates direction and steepness of the graph.</li> <li>Intercepts, both x- and y-, anchor the line on the axes and aid in quick graphing.</li> <li>Multiple forms exist for representing lines:<ul> <li>Slope-intercept:y = mx + b,emphasizesslopeandy−intercept.</li><li>Point−slope:, emphasizes slope and y-intercept.</li> <li>Point-slope:y - y1 = m(x - x1),usefulwhenapointonthelineisknown.</li><li>Standardform:, useful when a point on the line is known.</li> <li>Standard form:A x + B y = C,usefulforcertainalgebraicmanipulationsandsystems.</li></ul></li><li>Convertingbetweenformsinvolvesalgebraicrearrangement,withthesloperemindingthat, useful for certain algebraic manipulations and systems.</li></ul></li> <li>Converting between forms involves algebraic rearrangement, with the slope reminding thatm = -\frac{A}{B}wheninstandardform.</li><li>Real−worldrelevance:linearmodelsdescribeconstant−ratechangessuchasdepreciation,simpleinterest,andbasiccost−revenuerelationships.</li><li>Ethical/practicalnote:choosingtherightmodelandunderstandingitslimitsisessential;linearmodelsassumeconstantrateofchangewhichmaynotholdforlonghorizonsornonlinearphenomena.</li></ul><h3id="quickpracticerecap">QuickPracticeRecap</h3><ul><li>Slopebetweentwopoints:when in standard form.</li> <li>Real-world relevance: linear models describe constant-rate changes such as depreciation, simple interest, and basic cost-revenue relationships.</li> <li>Ethical/practical note: choosing the right model and understanding its limits is essential; linear models assume constant rate of change which may not hold for long horizons or nonlinear phenomena.</li> </ul> <h3 id="quickpracticerecap">Quick Practice Recap</h3> <ul> <li>Slope between two points:m = \frac{y2 - y1}{x2 - x1}.</li><li>Horizontalline:.</li> <li>Horizontal line:y = k\Rightarrow m = 0.</li><li>Verticalline:.</li> <li>Vertical line:x = k\Rightarrow m\text{ undefined}.</li><li>X−interceptofline.</li> <li>X-intercept of liney = mx + b:sety=0,solveforx:: set y = 0, solve for x:0 = mx + b \Rightarrow x = -\frac{b}{m}(whenm≠0).</li><li>Standardformtoslope−intercept:(when m ≠ 0).</li> <li>Standard form to slope-intercept:y = -\frac{A}{B}x + \frac{C}{B}forforA x + B y = C.</li><li>Depreciationmodelexample:.</li> <li>Depreciation model example:V(t) = -1400 t + 20000.$$