# Copyright © 2014, 2010, 2007 Pearson Education, Inc. 1 1 Chapter 9 Quadratic Equations and Functions.

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Copyright © 2014, 2010, 2007 Pearson Education, Inc. 4 Axis of Symmetry: A line that divides a graph into two symmetrical halves. Vertex of a Parabola: The lowest point on a parabola that opens up or highest point on a parabola that opens down.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 5 Example Solution xy(x, y) 0 1 –1 2 –2 0228802288 (0, 0) (1, 2) (–1, 2) (2, 8) (–2, 8) Graph. The graph opens up because a is positive.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 6 Example Graph. Solution xy(x, y) 0 1 –1 2 –2 0 –3 –12 (0, 0) (1, –3) (–1, –3) (2, –12) (–2, –12) The graph opens down because a is negative.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 7 As the |a| increases, the parabolas appear narrower.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 8 Conclusion Given a function in the form f(x) = ax 2, the axis of symmetry is x = 0 (the y-axis) and the vertex is (0, 0). Also the greater the absolute value of a, the narrower the parabola appears (or, the smaller the absolute value of a, the wider the parabola appears).

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 9 Example Graph g(x) = 2x 2 – 3 and h(x) = 2x 2 + 3. Solution We will graph both functions on the same grid. The vertex of the basic function f(x) = 2x 2 is the origin (0, 0). The vertex of g(x) = 2x 2 – 3 is shifted down 3 units and the vertex of h(x) = 2x 2 + 3 is shifted up 3 units from the origin.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 10 Conclusion Given a function in the form f(x) = ax 2 + k, if k > 0, then the graph of f(x) = ax 2 is shifted k units up from the origin. If k < 0, then the graph of f(x) = ax 2 is shifted k units down from the origin. The new position of the vertex is (0, k). The axis of symmetry is x = 0 (the y-axis).

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 11 Example Graph: f(x) = 3(x – 3) 2 and n(x) = 3(x + 3) 2 Solution The vertex of the basic function f(x) = 3x 2 is (0, 0) and the axis of symmetry is x = 0. The graph of f(x) = 3(x – 3) 2 has the same shape, only the vertex and axis of symmetry are shifted right 3 units from the origin along the x-axis so that they are (3, 0) and x = 3. Similarly, the vertex of n(x) = 3(x + 3) 2 is shifted 3 units to the left.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 12 Conclusion Given a function in the form f(x) = a(x – h) 2, if h > 0, then the graph of f(x) = ax 2 is shifted h units right from the origin. If h < 0, then the graph of f(x) = ax 2 is shifted |h| units left from the origin. The new position of the vertex is (h, 0). The axis of symmetry is x = h.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 13 Parabola with Vertex (h, k) The graph of a function in the form f(x) = a(x – h) 2 + k is a parabola with vertex at (h, k). The equation of the axis of symmetry is x = h. The parabola opens upward if a > 0 and downward if a < 0. The larger the |a|, the narrower the graph. (h, k) a > 0 a < 0 x = h

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 14 Example Given determine whether the graph opens up or down, find the vertex and axis of symmetry, and draw the graph. Find the domain and range. Solution a = 2, h = 2, and k =  3. a is positive the parabola opens upwards. The vertex is at (2,  3) and the axis of symmetry is x = 2.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 15 continued Domain: {x|x is a real number} or (−∞,∞ ) Range: {y|y  −3} or [−3,∞ )

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 16 Example Write f(x) = x 2 – 6x + 8 in the form f(x) = a(x – h) 2 + k. Then determine whether the graph opens upwards or downwards, find the vertex and axis of symmetry, and draw the graph. Find the domain and range. Solution Complete the square. y = x 2 – 6x + 8 y – 8 = x 2 – 6x y – 8 + 9 = x 2 – 6x + 9 y + 1 = (x – 3) 2 y = (x – 3) 2 – 1 Since a = 1 and 1 is positive, the graph opens upwards. The vertex is (3,  1). The axis of symmetry is x = 3.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 17 continued y = x 2 – 6x + 8 xy(x, y) 0234602346 8 0 –1 0 8 (0, 8) (2, 0) (3, –1) (4, 0) (6, 8) y-intercept x-intercept vertex Domain: {x|x is a real number} or (−∞,∞ ) Range: {y|y  −1} or [−1,∞ )

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 18 Finding the Vertex of a Quadratic Function in the Form f(x) = ax 2 + bx + c Given an equation in the form y = ax 2 + bx + c, to determine the vertex, 1. Find the x-coordinate using the formula 2. Find the y-coordinate by evaluating

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 19 Example For the function f(x) = 3x 2 – 12x + 4, find the coordinates of the vertex. Solution The vertex is (2,  8).