1 / 25

Chapter 2 Linear Functions and Equations

Chapter 2 Linear Functions and Equations. Equations of Lines. 2.1. Write the point-slope and slope-intercept forms Find the intercepts of a line Write equations for horizontal, parallel, and perpendicular lines Model data with lines and linear functions (optional)

bobbye
Download Presentation

Chapter 2 Linear Functions and Equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 2 Linear Functions and Equations

  2. Equations of Lines 2.1 Write the point-slope and slope-intercept forms Find the intercepts of a line Write equations for horizontal, parallel, and perpendicular lines Model data with lines and linear functions (optional) Use linear regression to model data (optional)

  3. Point-Slope Form of the Equation of a Line The line with slope m passing through the point (x1, y1) has equation y = m(xx1) + y1 or y y1 = m(xx1), The point-slope form of the equation of a line.

  4. Find an equation of the line passing through the points (–2, –3) and (1, 3). Plot the points and graph the line by hand. Example: Determining a point-slope form Solution Calculate the slope:

  5. Example: Determining a point slope form Substitute (1, 3) for (x1, y1) and 2 for m y = m(x –x1) + y1 y = 2(x – 1) + 3 Or substitute (–2, –3) for (x1, y1) and 2 for m y = m(x –x1) + y1 y = 2(x – (–2)) + (–3) y = 2(x + 2) – 3

  6. Solution (continued) Example: Determining a point-slope form Here’s the graph:

  7. Slope-Intercept Form of the Equation of a Line • The line with slope m and y-interceptb is given by • y = mx+ b, • the slope-intercept form of the equation of a line.

  8. Find the slope-intercept form for the line passing through the points (–2, 1) and (2, 3). Example: Finding slope-intercept form Solution Determine m and b in the form y = mx + b Substitute either point to find b, use (2, 3).

  9. Finding Intercepts • To find any x-intercepts, let y = 0 in the equation and solve for x. • To find any y-intercepts, let x = 0 in the equation and solve for y. • To find any x-intercepts, let y = 0 in the equation and solve for x. • To find any y-intercepts, let x = 0 in the equation and solve for y.

  10. Locate the x- and y-intercepts for the line whose equation is 4x + 3y = 6. Use the intercepts to graph the equation. Example: Finding Intercepts Solution To find the x-intercept, let y = 0, solve for x: The x-intercept is 1.5

  11. To find the y-intercept, let x = 0, solve for y: Example: Finding Intercepts The y-intercept is 2. The graph passes through the points (1.5, 0) and (0, 2).

  12. Graph of a constant function f Formula: f (x) = b Horizontal line with slope 0 and y-intercept b. Horizontal Lines (-3, 3) (3, 3) Note that regardless of the value of x, the value of y is always 3.

  13. Vertical Lines Cannot be represented by a function Slope is undefined Equation is: x = k • Note that regardless of the value of y, the value of x is always 3. • Equation is x = 3 (or x + 0y = 3) • Equation of a vertical line is x = k where k is the x-intercept.

  14. Equations of Horizontal and Vertical Lines • An equation of the horizontal line with y-intercept b is y = b. • An equation of the vertical line with x-intercept k is x = k.

  15. Parallel Lines • Two lines with slopes m1 and m2, neither of which is vertical, are parallel if and only if their slopes are equal; that is, m1 = m2.

  16. Perpendicular Lines • Two lines with nonzero slopes m1 and m2,, are perpendicular if and only if their slopes have a product of –1; that is, m1m2 = –1.

  17. Perpendicular Lines For perpendicular lines, m1 and m2, are negative reciprocals.

  18. Example: Finding perpendicular lines Find the slope-intercept form of the line perpendicular to passing through the point (–2, 1) . Graph the lines. Solution The line has slope The negative reciprocal is Use the point-slope form of the line . . .

  19. Example: Finding perpendicular lines

  20. Example: Modeling investments for cloud computing The table lists the investments in billions of dollars for cloud computing for selected years. (a) Make a scatterplot of the data. (b) Find a formula in point-slope form for a linear function f that models the data. (c) Graph the data and y = f(x) in the same xy-plane. (d) Interpret the slope of the graph of y = f(x). (e) Estimate the investment in cloud computing in 2014. Does your answer involve interpolation or extrapolation?

  21. Solution Example: Modeling investments for cloud computing (a) Make a scatterplot of the data.

  22. Example: Modeling investments for cloud computing (b) First data point (2005, 26); last data point (2009, 374)

  23. Example: Modeling investments for cloud computing (c)

  24. Example: Modeling investments for cloud computing (d) Interpret the slope of the graph of y = f(x). Slope 87 indicates that investments increased, on average, by $87 billion per year between 2005 and 2009.

  25. Example: Modeling investments for cloud computing (e) Estimate the investment in cloud computing in 2014. Does your answer involve interpolation or extrapolation? To estimate the investments in 2014, we can evaluate f(2014). This model predicts an $809 billion investment in cloud computing during 2014. This result involves extrapolation because 2014 is “outside” of 2005 and 2009.

More Related