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§ 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 .

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§ 4.2

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  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

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