Name: Date: Algebra 2 Finding the Vertex Vocabulary and basic facts The graph of any quadratic function f(x) = ax2 + bx + c is called a parabola. The graph will have one of two shapes, and the a value tells which shape it will be. graph shape if a is positive graph shape if a is negative Every parabola has a special point called the vertex. It’s the lowest or highest point. Every parabola is symmetric across a vertical line called the axis of symmetry. The vertex is always on this line. The line’s equation is x = [the x-coordinate of the vertex]. Because of symmetry, when a quadratic function has two zeros, the vertex and the axis of symmetry are midway between the zeros. (In the example, the zeros are x = 3 and x = 7; the vertex and axis are at x = 5.) Three ways to find the vertex Here are three methods for finding the coordinates of the vertex, each covered by a part of today’s assignment: (1) Calculator (2) x-coordinate of the vertex is the average of the zeros. (3) formula for x-coordinate of the vertex: x = -b . 2a Name: Date: Algebra 2 You try it 1. For each function, use your calculator to decide whether the vertex is a maximum or a minimum, find the coordinates of the vertex, and write an equation for the axis of symmetry. a. f(x) = 2x2 + 4x + 7 maximum or minimum? vertex: axis of symmetry: x-intercept: y-intercept: b. f(x) = –3x2 + 6x maximum or minimum? vertex: axis of symmetry: x-intercept: y-intercept: c. f(x) = –x2 + 4x + 10 maximum or minimum? vertex: axis of symmetry: x-intercept: y-intercept: 2. Suppose that when a ball thrown from a cliff, its path is given by the equation: h(x) = –x2 + 10x + 24. What is the maximum height reached by this ball? 3. For the quadratic function f(x) = x2 + 3x – 24, make the table and graph on your calculator. Then, use the calculator to find the zeros, vertex, and y-intercept. zeros: vertex: y-intercept: Name: Date: Algebra 2 Method 2: Finding the vertex by averaging the zeros Here is how to use a quadratic’s zeros to find the coordinates of the vertex: First find the zeros by any method (such as factoring or the Quadratic Formula). Find the x-coordinate of the vertex by averaging the zeros (add the zeros then divide by 2). Then, you can evaluate f(x) to find out the y-coordinate of the vertex. Example: Find the vertex and the axis of symmetry of f(x) = x2 + 2x – 35. Solution: Factor the function: f(x) = (x – 5)(x + 7). Then find the zeros: x = 5, x = –7 x-coordinate of vertex: x= y-coordinate of vertex: y = f(–1) = (–1)2 + 2(–1) – 35 = 1 – 2 – 35 = –36. 5 + ( -7 ) = –1. 2 Answer: The vertex is (–1, –36). The axis of symmetry is the line x = –1. You try it 4. For each function, use factoring to find the zeros. Then, find the coordinates of the vertex and an equation for the axis of symmetry and zeros. a. f(x) = x2 – 4x – 60 b. f(x) = 6x2 – 5x + 1 c. f(x) = –x2 + 3x + 70 Name: Date: Algebra 2 5. Find the zeros of these functions using factoring, then find the coordinates of the vertex. a. f(x) = 5x2 + 20x + 15 b. f(x) = 3x2 + 6 x c. f(x) = 4x2 – 9 6. Check your answers to all parts of problems 4 and 5 by finding the vertex of each function on your calculator. Record each vertex from the calculator here, and confirm agreement with your previous answers. Fix any mistakes that you find. 4a. 5a. 4b. 5b. 4c. 5c. Name: Date: Algebra 2 Method 3: Finding the vertex using formula x = -b 2a Here is how to find a quadratic’s vertex using a formula. -b 2a Then, you can evaluate f(x) to find out the y-coordinate of the vertex. The x-coordinate of a parabola’s vertex is always x = Example: Find the vertex and the axis of symmetry of f(x) = –3x2 + 12x + 4. Solution: x= - 12 -b - 12 = = = 2. 2 × ( -3) 2a -6 y = f(2) = –3 · 22 + 12 · 2 + 4 = –12 + 24 + 4 = 16. Answer: The vertex is (2, 16). The axis of symmetry is the line x = 2. You try it 7. Using the formula shown above, find the vertex and the axis of symmetry and intercept with the coordinate axis for each of these functions. a. f(x) = 5x2 – 20x + 15 b. f(x) = 3x2 + 8x + 6 c. f(x) = x2 – 7 Name: Date: 8. For the following quadratic functions, find the vertex, zeros, axis of symmetry and y intercept. Use your information to sketch a graph of the function. a. f ( x) = x2 - x - 6 Vertex:__________________ Zeros:___________________ Axis of symmetry:_________ Y-intercept:______________ b. f (x) = 6 x - x2 Hint: First rearrange the terms into standard form. Vertex:__________________ Zeros:___________________ Axis of symmetry:_________ Y-intercept:______________ Algebra 2 Name: Date: Algebra 2 Application: Giant Parabola 9. In St. Louis there is a large monument called the Gateway Arch that is shaped like a parabola. Specifically this function gives the shape of the arch is f(x) = –0.0089x2 + 4.737x. The x-axis represents the ground, and both variables are measured in feet. Gateway Arch photo: www.arrakeen.ch Why averaging the zeros doesn’t always work: 10. In problem 7b, you found the vertex of f(x) = 3x2 + 8x + 6 using the formula method. Explain why the method of averaging the zeros wouldn’t have worked for this function. Hint: Look at the graph on your calculator. 11. The function f(x) = x2 + 8x + 16 has just one zero a. Factor f(x), then find the zero of f(x). b. Using any method, find the coordinates of the vertex of f(x).