1 / 20

§ 4.2

§ 4.2. Graphing Linear Equations Using Intercepts. Linear Functions. All equations of the form Ax + By = C are straight lines when graphed, as long as A and B are not both zero. Such an equation is called the standard form of the equation of the line .

eadoin
Download Presentation

§ 4.2

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. §4.2 Graphing Linear Equations Using Intercepts

  2. Linear Functions All equations of the form Ax + By = C are straight lines when graphed, as long as A and B are not both zero. Such an equation is called the standard form of the equation of the line. To graph equations of this form, two very important points are used – the intercepts. Blitzer, Introductory Algebra, 5e – Slide #2 Section 4.2

  3. x - intercept An x-intercept of a graph is the x-coordinate of a point where the graph intersects the x-axis. The y-coordinate of the x-intercept is always zero. The graph of y = 4x – 8 crosses the x-axis at (2, 0) and that point is the x-intercept. (2, 0) Blitzer, Introductory Algebra, 5e – Slide #3 Section 4.2

  4. y- intercept The y-intercept of a graph is the y-coordinate of a point where the graph intersects the y-axis. The x-coordinate of the y-intercept is always zero. The graph of y = 3x + 4 crosses the y-axis at (0, 4) and that point is the y-intercept. (0, 4) Blitzer, Introductory Algebra, 5e – Slide #4 Section 4.2

  5. x- and y-Intercepts Blitzer, Introductory Algebra, 5e – Slide #5 Section 4.2

  6. Graphing Using Intercepts EXAMPLE Graph the equation using intercepts. SOLUTION 1) Find the x-intercept. Let y = 0 and then solve for x. Replace y with 0 Multiply and simplify Divide by -2 The x-intercept is -6, so the line passes through (-6,0). Blitzer, Introductory Algebra, 5e – Slide #6 Section 4.2

  7. Graphing Using Intercepts CONTINUED 2) Find the y-intercept. Let x = 0 and then solve for y. Replace x with 0 Multiply and simplify Divide by 4 The y-intercept is 3, so the line passes through (0,3). Blitzer, Introductory Algebra, 5e – Slide #7 Section 4.2

  8. Graphing Using Intercepts CONTINUED 3) Find a checkpoint, a third ordered-pair solution. For our checkpoint, we will let x = 1 (because x = 1 is not the x-intercept) and find the corresponding value for y. Replace x with 1 Multiply Add 2 to both sides Divide by 4 and simplify The checkpoint is the ordered pair (1,3.5). Blitzer, Introductory Algebra, 5e – Slide #8 Section 4.2

  9. Graphing Using Intercepts CONTINUED 4) Graph the equation by drawing a line through the three points. The three points in the figure below lie along the same line. Drawing a line through the three points results in the graph of . (0,3) (1,3.5) (-6,0) Blitzer, Introductory Algebra, 5e – Slide #9 Section 4.2

  10. Graphing Using Intercepts ExampleUse intercepts for graphing 4x + 3y = 12 • Find the x-intercept. Let y = 0 and solve for x. 4x + 3(0) = 12 4x = 12 x = 3 The x-intercept is 3. The line passes through (3, 0). • Find the y-intercept. Let x = 0 and solve for y. 4(0) + 3y = 12 3y = 12 y = 4 The y-intercept is 4. The line passes through (0, 4). Blitzer, Introductory Algebra, 5e – Slide #10 Section 4.2

  11. Graphing Using Intercepts Example , continued: Graph 4x + 3y = 12. • Find a checkpoint, a third ordered-pair solution. Let x = 2 4(2) + 3y = 12 8 + 3y = 12 3y = 4 y = 4/3 ~ 1.33 A checkpoint is the ordered pair (2, 1.33) Blitzer, Introductory Algebra, 5e – Slide #11 Section 4.2

  12. Graphing Using Intercepts Example , continued: Graph 4x + 3y = 12. • Graph the equation by drawing a line through the intercepts and checkpoint. (0, 3) (2, 1.33) (4, 0) Blitzer, Introductory Algebra, 5e – Slide #12 Section 4.2

  13. Horizontal and Vertical Lines Blitzer, Introductory Algebra, 5e – Slide #13 Section 4.2

  14. Horizontal and Vertical Lines EXAMPLE Graph the linear equation: y = -4. SOLUTION All ordered pairs that are solutions of y = -4 have a value of y that is always -4. Any value can be used for x. Let’s select three of the possible values for x: -3, 1, 6: (1,-4) (-3,-4) (6,-4) Upon plotting the three resultant points and connecting the points with a line, the graph to the right is the solution. Blitzer, Introductory Algebra, 5e – Slide #14 Section 4.2

  15. Horizontal and Vertical Lines EXAMPLE Graph the linear equation: x = 5. SOLUTION All ordered pairs that are solutions of x = 5 have a value of x that is always 5. Any value can be used for y. Let’s select three of the possible values for y: -3, 1, 5: (5,5) (5,1) (5,-3) Upon plotting the three resultant points and connecting the points with a line, the graph to the right is the solution. Blitzer, Introductory Algebra, 5e – Slide #15 Section 4.2

  16. Horizontal Lines Remember that the graph of y = b is a horizontal line. The y-intercept is b. y = b y (0, b) x Blitzer, Introductory Algebra, 5e – Slide #16 Section 4.2

  17. Horizontal Lines Example: Consider the graph of y = 2. This is a horizontal line and the y-intercept is 2. y = 2 1 -3 -2 -11 2 3 y 2 (0, 2) x Blitzer, Introductory Algebra, 5e – Slide #17 Section 4.2

  18. Vertical Lines Remember that the graph of x = a is a vertical line. The x-intercept is a. x = a y x (a, 0) Blitzer, Introductory Algebra, 5e – Slide #18 Section 4.2

  19. Vertical Lines Consider the graph of x = -3. The graph is a vertical line. The x-intercept is -3. x = -3 y x -3 Blitzer, Introductory Algebra, 5e – Slide #19 Section 4.2

  20. Graphing Using Intercepts – In Summary • Find the x-intercept. Let y = 0 and solve for x. • Find the y-intercept. Let x = 0 and solve for y. • Find a checkpoint, a third ordered-pair solution. • Graph the equation by drawing a line through the three points. Blitzer, Introductory Algebra, 5e – Slide #20 Section 4.2

More Related