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Graphing Parabolas

Learn how to graph parabolas using the vertex, axis of symmetry, and y-intercept with detailed examples and explanations. Explore various forms, coefficients, and transformations of parabolic equations. Practice identifying vertices and axes for different quadratic functions. Discover strategies for maximizing income in real-world scenarios.

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Graphing Parabolas

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  1. Graphing Parabolas Using the Vertex Axis of Symmetry & y-Intercept By: Jeffrey Bivin Lake Zurich High School jeff.bivin@lz95.org Last Updated: October 15, 2007

  2. Graphing Parabolas • With your graphing calculator, graph each of the following quadratic equations and identify the vertex and axis of symmetry. y = x2 + 4x - 7 y = 2x2 + 10x + 4 y = -3x2 + 5x + 9 Jeff Bivin -- LZHS

  3. Graph the following parabola x = -2 y = x2 + 4x - 7 vertex: axis of symmetry: (0, -7) (-2, -11) y-intercept: Jeff Bivin -- LZHS

  4. Graph the following parabola y = 2x2 + 10x + 4 vertex: axis of symmetry: y-intercept: Jeff Bivin -- LZHS

  5. Graph the following parabola y = -3x2 + 5x + 9 vertex: axis of symmetry: y-intercept: Jeff Bivin -- LZHS

  6. Graphing Parabolas • Now look at the coefficients of the equation and the value of the axis of symmetry – especially a and b • y = ax2+ bx + c y = x2 + 4x - 7 y = 2x2 + 10x + 4 y = -3x2 + 5x + 9 Jeff Bivin -- LZHS

  7. Graphing Parabolas y = ax2 + bx + c Axis of symmetry: Vertex: Jeff Bivin -- LZHS

  8. Graph the following parabola x = -2 re-visited y = x2 + 4x - 7 axis of symmetry: vertex: (0, -7) (-2, -11) y-intercept: Jeff Bivin -- LZHS

  9. Graph the following parabola re-visited y = 2x2 + 10x + 4 (0, 4) axis of symmetry: vertex: y-intercept: Jeff Bivin -- LZHS

  10. Graph the following parabola Why did this parabola open downward instead of upward as did the previous two? re-visited y = -3x2 + 5x + 9 axis of symmetry: vertex: y-intercept: Jeff Bivin -- LZHS

  11. Graph the following parabola y = x2 + 6x - 8 x = -3 Axis of symmetry: (0, -8) Vertex: (-3, -17) y-intercept: Jeff Bivin -- LZHS

  12. Graph the following parabola y = -2x2 + 7x + 12 (0, 12) Axis of symmetry: Vertex: y-intercept: Jeff Bivin -- LZHS

  13. Graphing Parabolas In Vertex Form Jeff Bivin -- LZHS

  14. Graphing Parabolas • With your graphing calculator, graph each of the following quadratic equations and identify the vertex and axis of symmetry. y = x2 vertex axis of sym. y = (x - 5)2 - 4 y = -3(x + 2)2 + 5 y = ⅜(x - 3)2 + 1 Jeff Bivin -- LZHS

  15. Graph the following parabola x = 5 y = (x - 5)2 - 4 (0, 21) axis of symmetry: vertex: (5, 4) y-intercept: Jeff Bivin -- LZHS

  16. Graph the following parabola (-2, 5) y = -3(x + 2)2 + 5 axis of symmetry: vertex: (0, -7) x = -2 y-intercept: Jeff Bivin -- LZHS

  17. Graph the following parabola x = 3 y = ⅜•(x - 3)2 - 1 axis of symmetry: vertex: y-intercept: (3, -1) Jeff Bivin -- LZHS

  18. Graphing Parabolas In Intercept Form Jeff Bivin -- LZHS

  19. Graph the following parabola x = 1 0 y = (x – 4)(x + 2) x-intercepts: (-2, 0) (4, 0) (0, -8) axis of symmetry: (1, -9) vertex: y-intercept: Jeff Bivin -- LZHS

  20. Graph the following parabola x = 5 0 y = (x - 1)(x - 9) (0, 9) x-intercepts: (1, 0) (9, 0) axis of symmetry: (5, -16) vertex: y-intercept: Jeff Bivin -- LZHS

  21. Graph the following parabola 0 y = -2(x + 1)(x - 5) (2, 18) (0, 10) x-intercepts: (-1, 0) (5, 0) axis of symmetry: x = 2 vertex: y-intercept: Jeff Bivin -- LZHS

  22. Convert to standard form y = -2(x + 1)(x - 5) y = -2(x2 – 5x + 1x – 5) y = -2(x2 – 4x – 5) y = -2x2 + 8x + 10 Jeff Bivin -- LZHS

  23. Now graph from standard form. y = -2x2 + 8x + 10 (2, 18) Axis of symmetry: (0, 10) Vertex: x = 2 y-intercept: Jeff Bivin -- LZHS

  24. A taxi service operates between two airports transporting 200 passengers a day. The charge is $15.00. The owner estimates that 10 passengers will be lost for each $2 increase in the fare. What charge would be most profitable for the service? What is the maximum income? VERTEX Income = Price ● Quantity Define the variable x = number of $2 price increases f(x) = ( 15 + 2x ) ( 200 – 10x ) 15 + 2x = 0 200 – 10x = 0 f(x) = income 2x = -15 200 = 10x Vertex is: So, price = (15 + 2x) = (15 + 2(6.25)) = 15 + 12.5 = $27.50 Maximumincome: 137.50 27.50 Jeff Bivin -- LZHS

  25. Alternative Method A taxi service operates between two airports transporting 200 passengers a day. The charge is $15.00. The owner estimates that 10 passengers will be lost for each $2 increase in the fare. What charge would be most profitable for the service? What is the maximum income? VERTEX Income = Price ● Quantity Define the variable x = number of $2 price increases f(x) = ( 15 + 2x ) ( 200 – 10x ) f(x) = 3000 – 150x + 400x – 20x2 f(x) = income f(x) = – 20x2 + 250x + 3000 f(6.25) = – 20(6.25)2 + 250(6.25) + 3000 f(6.25) = 3781.25 Vertex is: So, price = (15 + 2x) = (15 + 2(6.25)) = 15 + 12.5 = $27.50 Maximum income = f(x) = $3781.25 Jeff Bivin -- LZHS

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