<p>2.1 Equations of Lines</p><p>Slope of a Line (p.68) The slope or “steepness” of a line containing the points is the ratio of the change in y over the change in x. The “change in y” or The “change in x” or The slope, m = </p><p>Point-Slope Form of the Equation of a Line (p.69)</p><p>For a line with slope m and passing through the point (x1, y1), the Point-Slope Form of the equation of the line is y = m(x  x1) + y1 or</p><p> y  y1 = m(x  x1)</p><p>Example 1 (p.69): Determine the point-slope form for the equation of the line passing through the points (–2, –3) and (1, 3). Then plot the points and graph the line by hand. Solution: Calculate the slope: m = = 2 </p><p>Now, substitute the coordinates of either point for (x1, y1) and 2 for m</p><p>Substituting (1, 3) for (x1, y1) and 2 for m</p><p> y = m(x –x1) + y1 y = 2(x – 1) + 3 This is the Point–Slope Form Continue solving for y, y = 2x – 2 + 3 y = 2x + 1</p><p>Substituting (–2, –3) for (x1, y1) and 2 for m</p><p> y = m(x –x1) + y1 y = 2(x – (–2)) + (–3) y = 2(x + 2) – 3 This is the Point–Slope Form</p><p>Continue solving for y, y = 2x + 4 – 3 y = 2x + 1 This is the same as for (1, 3).</p><p>Here’s the graph: Slope-Intercept Form of the Equation of a Line (p.70) For a line with slope m and y-intercept b, the Slope-Intercept Form of the equation of the line is y = mx + b </p><p>Example 3 (p.70): Find the slope-intercept form for the equation of the line passing through the points (–2, 1) and (2, 3). Solution: Calculate the slope: m = </p><p>Now use the coordinates of either point to find b. Using (2, 3), substitute 2 for x and 3 for y. </p><p>The equation is </p><p>Standard Form of the equation of a line (p.71) is the Standard Form for the equation of a line. a, b and c are constants with a and b not both zero. Examples of linear equations written in Standard Form:</p><p>Finding Intercepts (p.72) The x-intercept is the point where the line intersects the x-axis. The coordinates of any point on the x-axis are (#,0). To find the x-intercept, let y = 0 in the equation and solve for x.</p><p>The y-intercept is the point where the line intersects the y-axis. The coordinates of any point on the y-axis are (0,#). To find the y-intercept, let x = 0 in the equation and solve for y.</p><p>Example 5 (p.72): Find the x- and y-intercepts for the line whose equation is 4x + 3y = 6. Use the intercepts to graph the equation. Solution: To find the x-intercept, let y = 0 and solve for x.</p><p>The x-intercept is (1.5, 0)</p><p>To find the y-intercept, let x = 0 and solve for y.</p><p>The y-intercept is (0, 2)</p><p>Here’s the graph:</p><p>Horizontal Lines Graph of a constant function f Formula: f (x) = b Horizontal line has slop= 0 and y-intercept = b.</p><p>Note that regardless of the value of x, the value of y is always 3.</p><p>Vertical Lines Cannot be represented by a function Slope is undefined Equation is: x = k </p><p>Note that regardless of the value of y, the value of x is always 3. </p><p>Equation is x = 3 or x + 0y = 3 Parallel Lines (p.73)</p><p>Two distinct lines with slopes m1 and m2 and y-intercepts , neither of which is vertical, are parallel if and only if they have the same slope, m1 = m2, and different y-intercepts, </p><p>NOTE: The phrase “if and only if” (abbreviated “iff”) is used when two statements are mathematically equivalent.</p><p>1. If two distinct lines are parallel, then m1 = m2 and 2. If two distinct lines have equal slopes and different y-intercepts, then they are parallel.</p><p>Perpendicular Lines (p.74)</p><p>Two lines with nonzero slopes m1 and m2 are perpendicular if and only if the product of their slopes is –1. m1m2 = –1 </p><p>This implies that for perpendicular lines, m1 and m2 are negative reciprocals. Example 8 (p.74): Find the slope-intercept form of the line perpendicular to and passing through the point (– 2, 1). Graph the lines.</p><p>Solution: The line has slope The negative reciprocal is Use the slope-intercept form of a line. </p><p>The equation is </p><p>Here’s the graph:</p><p>Modeling Data: Interpolation and Extrapolation (pp.75 and 76) Interpolation occurs when we estimate values between given data points. (p.75) Extrapolation occurs when we estimate values outside the given data. (p.76)</p><p>Interpolation is usually more accurate than extrapolation. (p.76)</p><p>Example 10 (p.76): Modeling Investments for Cloud Computing The table lists the investments in billions of dollars for cloud computing for selected years.</p><p>Year 2005 2006 2007 2008 2009 Investments 26 113 195 299 374 a) Make a scatterplot of the data. </p><p> b) Find a linear function in point-slope form. First data point (2005, 26) Last data point (2009, 374) Find the slope. </p><p>Equation is c) Graph the data and y = f(x) in the same xy-plane. d) Interpret the slope of the graph of y = f(x).</p><p>The positive slope of 87 indicates that investments increased an average of $87 billion per year from 2005 to 2009. e) Estimate the investment in cloud computing in 2014. Does your answer involve interpolation or extrapolation?</p><p>To estimate the investments in 2014, we can evaluate for x = 2014. y = 87(2014-2005) + 26 = 809 This model predicts an $809 billion investment in cloud computing during 2014. This result involves extrapolation because 2014 is “outside” of the interval from 2005 to 2009.</p>