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Writing Equivalent Forms of Quadratic Functions Adapted from Walch Education.

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Presentation on theme: "Writing Equivalent Forms of Quadratic Functions Adapted from Walch Education."— Presentation transcript:

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2 Writing Equivalent Forms of Quadratic Functions Adapted from Walch Education

3 Recall… The standard form, or general form, of a quadratic function is written as f(x) = ax 2 + bx + c, where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant term. The process of completing the square can be used to transform a quadratic equation from standard form to vertex form, f(x) = a(x – h) 2 + k. Vertex form can be used to identify the key features of a function’s graph : Writing Equivalent Forms of Quadratic Functions

4 And…  The vertex of a parabola is the point where the graph changes from increasing to decreasing, or vice versa. In vertex form, the extremum of the graph of a quadratic equation is easily identified using the vertex, (h, k). If a < 0, the function achieves a maximum, where k is the y- coordinate of the maximum and h is the x-coordinate of the maximum. If a > 0, the function has a minimum, where k is the y- coordinate of the minimum and h is the x-coordinate of the minimum : Writing Equivalent Forms of Quadratic Functions

5 Don’t forget, Because the axis of symmetry goes through the vertex, the axis of symmetry is easily identified from vertex form as x = h. The process of factoring can be used to transform a quadratic equation in standard form to factored form, f(x) = a(x – r)(x – s). The zeros of a function are the x-values where the function value is 0. Setting the factored form equal to 0, 0 = a(x – r)(x – s), the zeros are easily identified as r and s : Writing Equivalent Forms of Quadratic Functions

6 One more thing… The axis of symmetry is easily identified from the factored form as the axis of symmetry occurs at the midpoint between the zeros. Therefore, the axis of symmetry is : Writing Equivalent Forms of Quadratic Functions

7 Practice # 1 Suppose that the flight of a launched bottle rocket can be modeled by the equation y = –x 2 + 6x, where y measures the rocket’s height above the ground in meters and x represents the rocket’s horizontal distance in meters from the launching spot at x = 0. How far does the bottle rocket travel in the horizontal direction from launch to landing? What is the maximum height the bottle rocket reaches? How far has the bottle rocket traveled horizontally when it reaches its maximum height? : Writing Equivalent Forms of Quadratic Functions

8 Identify the zeros of the function  In the original equation, y represents the height of the bottle rocket. At launch and landing, the height of the bottle rocket is 0.  Write the original equation in factored form. Set it equal to 0 to identify the zeros of the function : Writing Equivalent Forms of Quadratic Functions

9 Continued,  y = –x 2 + 6xOriginal equation  0 = –x 2 + 6xSet the equation equal to 0.  0 = –(x 2 – 6x) Factor out –1.  0 = –x(x – 6)Factor the binomial.  Solve for x by setting each factor equal to 0.  –x = 0 or x – 6 = 0  x = 0 or x = : Writing Equivalent Forms of Quadratic Functions

10 Continued,  The x-intercepts are at x = 0 and x = 6. Find the distance between the two points to determine how far the bottle rocket travels in the horizontal direction. 6 – 0 = 6  The bottle rocket travels 6 meters in the horizontal direction from launch to landing : Writing Equivalent Forms of Quadratic Functions

11 Determine the maximum height of the bottle rocket  The maximum height occurs at the vertex.  Write the equation in vertex form by completing the square : Writing Equivalent Forms of Quadratic Functions

12 Continued, : Writing Equivalent Forms of Quadratic Functions y = –x 2 + 6xOriginal equation y = –(x 2 – 6x) Factor out the common factor, –1, from the variable terms. y = –(x 2 – 6x + 9) + 9 Add and subtract the square of of the x-term. Be sure to multiply the subtracted term by a, –1. y = –(x – 3) Write the trinomial as a binomial squared and simplify the constant term.

13 Continued,  The vertex form is y = –(x – 3) The vertex is (3, 9). The maximum value is the y-coordinate of the vertex, 9.  The bottle rocket reaches a maximum height of 9 meters : Writing Equivalent Forms of Quadratic Functions

14 Determine the horizontal distance from the launch point to the maximum height of the bottle rocket.  We know that the bottle rocket is launched from the point (0, 0) and reaches a maximum height at (3, 9). Subtract the x-values of the two points to find the distance traveled horizontally. 3 – 0 = : Writing Equivalent Forms of Quadratic Functions

15 Continued, AAnother method is to take the total distance traveled horizontally from launch to landing and divide it by 2 to find the same answer. This is because the maximum value occurs halfway between the zeros of the function. The bottle rocket has traveled 3 meters horizontally when it reaches its maximum height : Writing Equivalent Forms of Quadratic Functions

16 Thanks for Watching!!! Ms. Dambreville


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