 # Adapted from Walch Education  The standard form of a quadratic function is f ( x ) = ax 2 + bx + c, where a is the coefficient of the quadratic term,

## Presentation on theme: "Adapted from Walch Education  The standard form of a quadratic function is f ( x ) = ax 2 + bx + c, where a is the coefficient of the quadratic term,"— Presentation transcript:

 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

  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

 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

  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

 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

 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

 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.

  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.

 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

 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

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

 Thanks for Watching! Ms. Dambreville

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