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Graphing the quadratic y – k = a(x – h) 2

Graphing the quadratic y – k = a(x – h) 2. Algebra IIB Mrs. Crespo 2012-2013. The Basic Graph of a quadratic function. (0,0). parabola. THE “ a ” in y – k = a (x – h ) 2. IF a > 0, the parabola opens upward. IF a < 0, the parabola opens downward. (0,0).

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Graphing the quadratic y – k = a(x – h) 2

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  1. Graphing the quadratic y – k = a(x – h)2 Algebra IIB Mrs. Crespo 2012-2013

  2. The Basic Graph of a quadratic function (0,0) parabola

  3. THE “a” in y – k = a(x – h)2 . • IF a > 0, the parabola opens upward. • IF a < 0, the parabola opens downward. (0,0)

  4. THE “k” in y – k = a(x – h)2 . • IF k > 0, the parabola moves up “k” units. • IF k < 0, the parabola moves down “k” units. (0,0)

  5. THE “h” in y – k = a(x – h)2 . • IF h > 0, the parabola moves right “h” units. • IF h < 0, the parabola moves left “h” units. (0,0)

  6. ALL TOGETHER AND MORE:y – k = a(x – h)2 • IF a > 0, the parabola opens upward. • IF a < 0, the parabola opens downward. • IF k > 0, the parabola moves up “k” units. • IF k < 0, the parabola moves down “k” units. • IF h > 0, the parabola moves right “h” units. • IF h < 0, the parabola moves left “h” units. To plot for now, we need: • vertex (h, k) • axis of symmetry x = h x = h V (h,k) V (h,k) x = h

  7. Graph y + 2 = (x + 3)2 x = -3 • a = 1 > 0, the parabola opens upward. • k = -2 < 0, the parabola moves down 2 units. • h = -3 < 0, the parabola moves left 3 units. • vertex (h, k) = (-3, -2) • axis of symmetry x = h is x = -3. V (-3,-2)

  8. Graph y - 3 = -(x + 1)2 • a = -1 < 0, the parabola opens downward. • k = 3 > 0, the parabola moves up 3 units. • h = -1 < 0, the parabola moves left 1 unit. • vertex (h, k) = (-1, 3) • axis of symmetry x = h is x = -1.

  9. Find an equation of the parabola • A parabola has vertex (-1, -2) and contains the point (2, -5). SOLUTION: Plug in vertex (h, k) on y – k = a(x – h)2 So, y – (-2)= a(x – (-1))2 Then, y +2 = a(x + 1)2 Solve for a with point (2, -5) The equation of the parabola is: y +2 = (x + 1)2 = a -5 +2 = a(2 + 1)2 -3 = a(3)2 = a -3 = 9a

  10. Find an equation of the parabola • A parabola has vertex (2, -3) and y-intercept 9. SOLUTION: Plug in vertex (h, k) on y – k = a(x – h)2 So, y – (-3)= a(x – 2)2 Then, y +3 = a(x -2)2 Solve for a with point (0, 9) The equation of the parabola is: y +3 = 3 (x - 2)2 = a 9 +3 = a(0 - 2)2 12 = a(-2)2 3 = a 12 = 4a

  11. Quest 7-5 • Graph y + 2 = (x – 1) 2 • Find an equation of the parabola with vertex (4,5) and contains (5,3) • Find an equation of the parabola with vertex (-1, -2) with a = -2.

  12. Homework 7-5 • Page 331 page 331 1-12 odd (just draw a reasonable graph without finding the intercepts at this time) page 332 19-21 all Algebra

  13. Acknowledgement PowerPoint by Mrs. Crespo for Algebra IIB 2012-2013 McDougall Little Algebra and Trigonometry Book 2 by Brown, Dolciani, Sorgenfrey, Kane 2011

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