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Published byGordon Maxwell Modified over 5 years ago

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Adapted from Walch Education

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The standard form of a quadratic function is 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 x -intercepts are the points at which the graph crosses the x -axis, and are written as ( x, 0). The y -intercept is the point at which the graph crosses the y -axis and is written as (0, y ). 5.3.1: Creating and Graphing Equations Using Standard Form 2 Key Concepts

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The axis of symmetry of a parabola is the line through the vertex of a parabola about which the parabola is symmetric. The axis of symmetry extends through the vertex of the graph. The vertex of a parabola is the point on a parabola where the graph changes direction, ( h, k ), where h is the x -coordinate and k is the y -coordinate. The equation of the axis of symmetry is or 5.3.1: Creating and Graphing Equations Using Standard Form 3 Axis of Symmetry

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To find the y -coordinate, substitute the value of x into the original function, The maximum is the largest y -value of a quadratic and the minimum is the smallest y -value. If a > 0, the parabola opens up and therefore has a minimum value. If a < 0, the parabola opens down and therefore has a maximum value. 5.3.1: Creating and Graphing Equations Using Standard Form 4 Key Concepts

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x -intercepts, y -intercept, where the function is increasing and decreasing, where the function is positive and negative, relative minimums and maximums, symmetries, and end behavior of the function. 5.3.1: Creating and Graphing Equations Using Standard Form 5 Key Features of a Quadratic

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h ( x ) = 2 x 2 – 11 x + 5 is a quadratic function. Determine the direction in which the function opens, the vertex, the axis of symmetry, the x -intercept(s), and the y - intercept. Use this information to sketch the graph. 5.3.1: Creating and Graphing Equations Using Standard Form 6 Practice # 1

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1.Determine whether the graph opens up or down. h ( x ) = 2 x 2 – 11 x + 5 is in standard form; therefore, a = 2. Since a > 0, the parabola opens up. 5.3.1: Creating and Graphing Equations Using Standard Form 7 Solution

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2.Find the vertex and the equation of the axis of symmetry. h ( x ) = 2 x 2 – 11 x + 5 is in standard form; therefore, a = 2 and b = –11. The vertex has an x -value of 2.75. 5.3.1: Creating and Graphing Equations Using Standard Form 8 Solution, continued Equation to determine the vertex Substitute 2 for a and –11 for b. x = 2.75Simplify.

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The y -value of the vertex is –10.125. The vertex is the point (2.75, –10.125). Since the axis of symmetry is the vertical line through the vertex, the equation of the axis of symmetry is x = 2.75. 9 Solution, continued h(x) = 2x 2 – 11x + 5 Original equation h(2.75) = 2(2.75) 2 – 11(2.75) + 5 Substitute 2.75 for x. h(2.75) = –10.125Simplify.

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3.Find the y -intercept. h ( x ) = 2 x 2 – 11 x + 5 is in standard form, so the y - intercept is the constant c, which is 5. The y -intercept is (0, 5). 5.3.1: Creating and Graphing Equations Using Standard Form 10 Solution, continued

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4.Find the x -intercepts, if any exist. The x -intercepts occur when y = 0. Substitute 0 for the output, h ( x ), and solve. Solved by factoring: h ( x ) = 2 x 2 – 11 x + 5 0 = 2 x 2 – 11 x + 5 0 = (2 x – 1)( x – 5) 0 = 2 x – 1 or 0 = x – 5 x = 0.5 or x = 5 The x -intercepts are (0.5, 0) and (5, 0). 5.3.1: Creating and Graphing Equations Using Standard Form 11 Solution, continued

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5.3.1: Creating and Graphing Equations Using Standard Form 12 Plot the points

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Thanks for Watching! Ms. Dambreville

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