# Drawing Graphs of Quadratic Functions

## Presentation on theme: "Drawing Graphs of Quadratic Functions"— Presentation transcript:

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. 22.0 Students use the quadratic formula or factoring techniques or 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

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.

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.

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 –2 2 4 6 y x –4 –6 (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…

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) –2 2 4 6 y x –4 –6 (ii) To find the y-intercept, put x = 0 into y = x2 – 3x This gives y = 2, so the y-intercept is at (0, 2). (iii) The x-coordinate of the vertex is always halfway between the x-intercepts. So the x-coordinate of the vertex is given by: x = = 3 2 1 + 2 Solution continues…

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) –2 2 4 6 y x –4 –6 And the y-coordinate of the vertex is: y = x2 – 3x + 2 3 2 1 4 – 3 × = – So the vertex is at , – . 3 2 1 4 Also, the parabola’s line of symmetry passes through the vertex. So, the line of symmetry is the line x = . 3 2 Solution continues…

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) –2 2 4 6 y x –4 –6 The next function is the same as in the previous example, only multiplied by –2. y = x2 – 3x + 2 The coefficient of x2 is negative this time, so the graph is concave down. Solution continues…

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) –2 2 4 6 y x –4 –6 (i) To find the x-intercepts of the graph of y = –2x2 + 6x – 4, you need to solve: –2x2 + 6x – 4 = 0 y = x2 – 3x + 2 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…

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) –2 2 4 6 y x –4 –6 (ii) Put x = 0 into y = –2x2 + 6x – 4 to find the y-intercept. y = x2 – 3x + 2 The y-intercept is (0, –4). Solution continues…

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) 3 2 –2 2 4 6 y x –4 –6 (iii) The vertex is at x = . y = x2 – 3x + 2 3 2 1 –2 × × – 4 = So the y-coordinate of the vertex is at: and, the line of symmetry is the line x = . 3 2 The coordinates of the vertex are , , 1 y = –2x2 + 6x – 4

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). –6 –4 –2 2 4 6 y x x-coordinate: [1 + (–1)] ÷ 2 = 0. y-coordinate: y = 0 – 1 = –1. So, the vertex is at (0, –1). Solution follows…

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). –6 –4 –2 2 4 6 y x x-coordinate: [3 + (–1)] ÷ 2 = 1. y-coordinate: y = (1 – 1)2 – 4 = –4. So, the vertex is at (1, –4). Solution follows…

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. –6 –4 –2 2 4 6 y x 1. y = x2 – 2x 1 2 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…

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. –6 –4 –2 2 4 6 y x 3. y = –4x2 – 4x + 3 3 4 x-intercepts: (– , 0) and ( , 0) y-intercept: (0, 3), vertex: (– , 4) 3 2 1 4. y = x2 – 4 x-intercepts: (–2, 0) and (2, 0) y-intercept: (0, –4) vertex: (0, –4) Solution follows…

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 –4 –2 2 4 6 8 10 y x 5. y = x2 + 4x + 4 6 5 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…

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. –6 –4 –2 2 4 6 y x 7. y = –9x2 – 6x + 3 7 x-intercepts: (–1 , 0) and ( , 0) y-intercept: (0, 3), vertex: (– , 4) 1 3 Solution follows…

Topic 7.4.2 Drawing Graphs of Quadratic Functions Independent Practice Describe the characteristics of quadratic graphs of the form y = ax2 + bx + c that 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…