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Topic 7.4.2

Drawing Graphs of Quadratic Functions. Topic 7.4.2. Topic 7.4.2. Drawing Graphs of Quadratic Functions. California Standards: 21.0 Students graph quadratic functions and know that their roots are the x -intercepts.

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Topic 7.4.2

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  1. Drawing Graphs of Quadratic Functions Topic 7.4.2

  2. Topic 7.4.2 Drawing Graphs of Quadratic Functions California Standards: 21.0 Students graph quadratic functions and know that their roots are the x-intercepts. 22.0 Students usethe quadratic formula or factoring techniquesor both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points. What it means for you: You’ll graph quadratic functions by finding their roots. • Key words: • quadratic • parabola • intercept • vertex • line of symmetry • root

  3. Topic 7.4.2 Drawing Graphs of Quadratic Functions In this Topic you’ll use methods for finding the intercepts and the vertex of a graph to draw graphs of quadratic functions.

  4. Topic 7.4.2 Drawing Graphs of Quadratic Functions Find the Roots of the Corresponding Equations In general, a good way to graph the function y = ax2 + bx + c is to find: (i) the x-intercepts (if there are any) — this involves solving a quadratic equation, (ii) the y-intercept — this involves setting x = 0, (iii) the vertex.

  5. y 6 4 2 x 0 –2 0 2 4 –2 –4 –6 Topic 7.4.2 Drawing Graphs of Quadratic Functions Example 1 Sketch the graphs of y = x2 – 3x + 2 and y = –2x2 + 6x – 4. Solution (i) To find the x-intercepts of the graph of y = x2 – 3x + 2, you need to solve: x2 – 3x + 2 = 0 This quadratic factors to give: (x – 1)(x – 2) = 0 Using the zero property, x = 1 or x = 2. So the x-intercepts are (1, 0) and (2, 0). Solution continues… Solution follows…

  6. y 6 3 1 + 2 4 2 2 2 x 0 –2 0 2 4 –2 –4 So the x-coordinate of the vertex is given by: x = = –6 Topic 7.4.2 Drawing Graphs of Quadratic Functions Example 1 Sketch the graphs of y = x2 – 3x + 2 and y = –2x2 + 6x – 4. Solution (continued) (ii) To find the y-intercept, put x = 0 into y = x2 – 3x + 2. This gives y = 2, so the y-intercept is at (0, 2). (iii) The x-coordinate of the vertexis always halfway between the x-intercepts. Solution continues…

  7. y 6 1 3 3 3 3 1 4 2 2 2 4 4 2 2 2 – 3 × + 2 = – So the vertex is at,– . x 0 –2 0 2 4 –2 –4 –6 So, the line of symmetry is the line x = . Topic 7.4.2 Drawing Graphs of Quadratic Functions Example 1 Sketch the graphs of y = x2 – 3x + 2 and y = –2x2 + 6x – 4. Solution (continued) y = x2 – 3x + 2 And the y-coordinate of the vertex is: Also, the parabola’s line of symmetry passes through the vertex. Solution continues…

  8. y 6 4 2 x 0 –2 0 2 4 –2 –4 –6 Topic 7.4.2 Drawing Graphs of Quadratic Functions Example 1 Sketch the graphs of y = x2 – 3x + 2 and y = –2x2 + 6x – 4. Solution (continued) y = x2 – 3x + 2 The next function is the same as in the previous example, only multiplied by –2. The coefficient of x2 is negative this time, so the graph is concave down. Solution continues…

  9. y 6 4 2 x 0 –2 0 2 4 –2 –4 –6 Topic 7.4.2 Drawing Graphs of Quadratic Functions Example 1 Sketch the graphs of y = x2 – 3x + 2 and y = –2x2 + 6x – 4. Solution (continued) y = x2 – 3x + 2 (i) To find the x-intercepts of the graph of y = –2x2 + 6x – 4, you need to solve: –2x2 + 6x – 4 = 0 This quadratic factors to give: –2(x – 1)(x – 2) = 0. Using the zero property,x = 1 or x = 2. This means the x-intercepts are at: (1, 0) and (2, 0). Solution continues…

  10. y 6 4 2 x 0 –2 0 2 4 –2 –4 –6 Topic 7.4.2 Drawing Graphs of Quadratic Functions Example 1 Sketch the graphs of y = x2 – 3x + 2 and y = –2x2 + 6x – 4. Solution (continued) y = x2 – 3x + 2 (ii) Put x = 0 into y = –2x2 + 6x – 4 to find the y-intercept. The y-intercept is (0, –4). Solution continues…

  11. y 6 3 3 1 1 3 3 3 4 So the y-coordinate of the vertexis at: 2 2 2 2 2 2 2 2 x 0 –2 0 2 4 –2 –4 2 The coordinates of thevertex are,, –2 × + 6 × – 4 = –6 and, the line of symmetry is the line x = . Topic 7.4.2 Drawing Graphs of Quadratic Functions Example 1 Sketch the graphs of y = x2 – 3x + 2 and y = –2x2 + 6x – 4. Solution (continued) y = x2 – 3x + 2 (iii) The vertexis at x = . y = –2x2 + 6x – 4

  12. y 6 4 2 x 0 –6 –4 –2 0 2 4 6 –2 –4 –6 Topic 7.4.2 Drawing Graphs of Quadratic Functions Guided Practice Exercises 1–4 are about the quadratic y = x2 – 1. 1. Find the x–intercepts (if there are any). 2. Find the y–intercepts (if there are any). 3. Find the vertex. 4. Using the vertex, x-intercepts, and y-intercepts, graph the quadratic. Let y = 0 and factor: 0 = (x – 1)(x + 1) so x = 1 or x = –1. So, the x-intercepts are (1, 0) and (–1, 0). When x = 0, y = 0 – 1 = – 1. So, the y-intercept is (0, –1). x-coordinate: [1 + (–1)] ÷ 2 = 0.y-coordinate: y = 0 – 1 = –1. So, the vertex is at (0, –1). Solution follows…

  13. y 6 4 2 x 0 –6 –4 –2 0 2 4 6 –2 –4 –6 Topic 7.4.2 Drawing Graphs of Quadratic Functions Guided Practice Exercises 5–8 are about the quadratic y = (x – 1)2 – 4. 5. Find the x–intercepts (if there are any). 6. Find the y–intercepts (if there are any). 7. Find the vertex. 8. Using the vertex, x-intercepts, and y-intercepts, graph the quadratic. Rearrange to form a standard quadratic:y = x2 – 2x – 3 Let y = 0 and factor: 0 = (x – 3)(x + 1) so x = 3 or x = –1. So, the x-intercepts are (3, 0) and (–1, 0). When x = 0, y = (0 – 1)2 – 4 = 1 – 4 = –3. So, the y-intercept is (0, –3). x-coordinate: [3 + (–1)] ÷ 2 = 1.y-coordinate: y = (1 – 1)2 – 4 = –4. So, the vertex is at (1, –4). Solution follows…

  14. y 6 4 2 x 0 –6 –4 –2 0 2 4 6 –2 –4 –6 Topic 7.4.2 Drawing Graphs of Quadratic Functions Independent Practice For each of the quadratics in Exercises 1–2, follow these steps: i) Find the x–intercepts (if any), ii) Find the y–intercepts (if any), iii) Find the vertex, iv) Using the vertex, x-intercepts, and y-intercepts, graph the quadratic. 1 2 1.y = x2 – 2x x-intercepts: (0, 0) and (2, 0)y-intercept: (0, 0)vertex: (1, –1) 2.y = x2 + 2x – 3 x-intercepts: (–3, 0) and (1, 0)y-intercept: (0, –3)vertex: (–1, –4) Solution follows…

  15. y 6 4 2 x 0 –6 –4 –2 0 2 4 6 3 1 1 –2 2 2 2 –4 x-intercepts: (– , 0) and ( , 0) y-intercept: (0, 3), vertex: (– , 4) –6 Topic 7.4.2 Drawing Graphs of Quadratic Functions Independent Practice For each of the quadratics in Exercises 3–4, follow these steps: i) Find the x–intercepts (if any), ii) Find the y–intercepts (if any), iii) Find the vertex, iv) Using the vertex, x-intercepts, and y-intercepts, graph the quadratic. 3 4 3.y = –4x2 – 4x + 3 4.y = x2 – 4 x-intercepts: (–2, 0) and (2, 0)y-intercept: (0, –4)vertex: (0, –4) Solution follows…

  16. y 10 8 6 4 2 x 0 –6 –4 –2 0 2 4 6 –2 Topic 7.4.2 Drawing Graphs of Quadratic Functions Independent Practice For each of the quadratics in Exercises 5–6, follow these steps: i) Find the x–intercepts (if any), ii) Find the y–intercepts (if any), iii) Find the vertex, iv) Using the vertex, x-intercepts, and y-intercepts, graph the quadratic. 6 5 5.y = x2 + 4x + 4 x-intercept: (–2, 0)y-intercept: (0, 4)vertex: (–2, 0) 6.y = –x2 + 4x + 5 x-intercepts: (5, 0) and (–1, 0)y-intercept: (0, 5)vertex: (2, 9) Solution follows…

  17. y 6 4 2 x 0 –6 –4 –2 0 2 4 6 1 1 –2 3 3 –4 x-intercepts: (–1, 0) and ( , 0) y-intercept: (0, 3), vertex: (– , 4) –6 Topic 7.4.2 Drawing Graphs of Quadratic Functions Independent Practice For the quadratics in Exercises 7, follow these steps: i) Find the x–intercepts (if any), ii) Find the y–intercepts (if any), iii) Find the vertex, iv) Using the vertex, x-intercepts, and y-intercepts, graph the quadratic. 7 7.y = –9x2 – 6x + 3 Solution follows…

  18. Topic 7.4.2 Drawing Graphs of Quadratic Functions Independent Practice Describe the characteristics of quadratic graphs of the form y = ax2 + bx + cthat have the following features, or say if they are not possible. 8. No x-intercepts 9. One x-intercept 10. Two x-intercepts 11. Three x-intercepts The graph is either concave up with the vertex above the x–axis, or concave down with the vertex below the x–axis. The vertex is the x–intercept. The graph is either concave up with the vertex below the x–axis, or concave down with the vertex above the x–axis. Not possible. Solution follows…

  19. Topic 7.4.2 Drawing Graphs of Quadratic Functions Independent Practice Describe the characteristics of quadratic graphs of the form y = ax2 + bx + cthat have the following features, or say if they are not possible. 12. No y-intercepts 13. One y-intercept 14. Two y-intercepts 15. Three y-intercepts Not possible. All quadratic equations of the form y=ax2 + bx + c will have one y–intercept. Not possible. Not possible. Solution follows…

  20. Topic 7.4.2 Drawing Graphs of Quadratic Functions Independent Practice 16. Which quadratic equation has the following features? Vertex (3, –4), x-intercepts (1, 0), (5, 0), and y-intercept (0, 5) y = (x – 3)2 – 4 or y = x2 – 6x + 5 17. Which quadratic equation has the following features? Vertex (0,16),x-intercepts(4,0),(–4,0),and y-intercept(0,16) y = –x2 + 16 Solution follows…

  21. Topic 7.4.2 Drawing Graphs of Quadratic Functions Round Up A quadratic function has the general form y = ax2 + bx + c (where a¹ 0). When you draw the graph of a quadratic, the value of a determines whether the parabola is concave up (u-shaped) or concave down (n-shaped), and how steep it is. Changing the value of c moves the graph in the direction of the y-axis. Note that if a = 0, the function becomes y = bx + c, which is a linear function whose graph is a straight line.

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