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1.2 Graphing Quadratic Functions In Vertex or Intercept Form

1.2 Graphing Quadratic Functions In Vertex or Intercept Form. Definitions 2 more forms for a quad. function Steps for graphing Vertex and Intercept Examples Changing between eqn. forms. Vertex Form Equation. y=a(x-h) 2 +k If a is positive, parabola opens up

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1.2 Graphing Quadratic Functions In Vertex or Intercept Form

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  1. 1.2 Graphing Quadratic Functions In Vertex or Intercept Form • Definitions • 2 more forms for a quad. function • Steps for graphing Vertex and Intercept • Examples • Changing between eqn. forms

  2. Vertex Form Equation y=a(x-h)2+k • If a is positive, parabola opens up If a is negative, parabola opens down. • The vertex is the point (h,k). • The axis of symmetry is the vertical line x=h. • Don’t forget about 1 point on either side of the vertex! (3 points total!)

  3. Example 2: Graphy=-½(x+3)2+4 y=a(x-h)2+k • a is negative (a = -½), so parabola opens down. • Vertex is (h,k) or (-3,4) • Axis of symmetry is the vertical line x = -3 • Table of values x y -1 2 -2 3.5 -3 4 -4 3.5 -5 2 Vertex (-3,4) (-4,3.5) (-2,3.5) (-5,2) (-1,2) x=-3

  4. Now you try one!y=a(x-h)2+k y=2(x-1)2+3 • Open up or down? • Vertex? • Axis of symmetry? • Table of values with 3 points?

  5. (-1, 11) (3,11) X = 1 (0,5) (2,5) (1,3)

  6. The Tacoma Narrows Bridge in Washington has two towers that each rise 307 feet above the roadway and are connected by suspension cables as shown. Each cable can be modeled by the function. 1 7000 y =(x – 1400)2 + 27 Civil Engineering where xand yare measured in feet. What is the distance dbetween the two towers ?

  7. SOLUTION The vertex of the parabola is (1400, 27). So, a cable’s lowest point is 1400 feet from the left tower shown above. Because the heights of the two towers are the same, the symmetry of the parabola implies that the vertex is also 1400 feet from the right tower. So, the distance between the two towers is d = 2 (1400) = 2800 feet.

  8. Intercept Form Equation y=a(x-p)(x-q) • The x-intercepts are the points (p,0) and (q,0). • The axis of symmetry is the vertical line x= • The x-coordinate of the vertex is • To find the y-coordinate of the vertex, plug the x-coord. into the equation and solve for y. • If a is positive, parabola opens up If a is negative, parabola opens down.

  9. Example 3: Graph y=-(x+2)(x-4)y=a(x-p)(x-q) • The axis of symmetry is the vertical line x=1 (from the x-coord. of the vertex) • Since a is negative, parabola opens down. • The x-intercepts are (-2,0) and (4,0) • To find the x-coord. of the vertex, use • To find the y-coord., plug 1 in for x. • Vertex (1,9) (1,9) (-2,0) (4,0) x=1

  10. Now you try one!y=a(x-p)(x-q) y=2(x-3)(x+1) • Open up or down? • X-intercepts? • Vertex? • Axis of symmetry?

  11. x=1 (-1,0) (3,0) (1,-8)

  12. The path of a placekicked football can be modeled by the function y = – 0.026x(x – 46) where xis the horizontal distance (in yards) and yis the corresponding height (in yards). Football a. How far is the football kicked ? b.What is the football’s maximum height ?

  13. SOLUTION a.Rewrite the function as y = – 0.026(x – 0)(x – 46). Because p = 0 and q = 46, you know the x - intercepts are 0 and 46. So, you can conclude that the football is kicked a distance of46 yards. b.To find the football’s maximum height, calculate the coordinates of the vertex.

  14. WHAT IF?In Example 4, what is the maximum height of the football if the football’s path can be modeled by the function y = –0.025x(x – 50)? SOLUTION a.Rewrite the function as y = – 0.025(x – 0) (x – 50). Because p = 0 and q = 50, you know the x - intercepts are 0 and 50. So, you can conclude that the football is kicked a distance of50 yards. b.To find the football’s maximum height, calculate the coordinates of the vertex.

  15. The maximum height is the y-coordinate of the vertex, or about 15.625 yards.

  16. Changing from vertex or intercepts form to standard form • The key is to FOIL! (first, outside, inside, last) • Ex: y=-(x+4)(x-9) Ex: y=3(x-1)2+8 =-(x2-9x+4x-36) =3(x-1)(x-1)+8 =-(x2-5x-36) =3(x2-x-x+1)+8 y=-x2+5x+36 =3(x2-2x+1)+8 =3x2-6x+3+8 y=3x2-6x+11

  17. Change from intercept form to standard form Write y = – 2 (x + 5) (x – 8) in standard form. Write original function. y = – 2 (x + 5) (x – 8) Multiply using FOIL. = – 2 (x 2 – 8x + 5x – 40) = – 2 (x 2 – 3x – 40) Combine like terms. = – 2x 2 + 6x + 80 Distributive property

  18. Change from vertex form to standard form Write f (x) = 4 (x – 1)2 + 9 in standard form. f (x) = 4(x – 1)2 + 9 Write original function. = 4(x – 1) (x – 1) + 9 Rewrite(x – 1)2. = 4(x 2 – x – x + 1) + 9 Multiply using FOIL. = 4(x 2 – 2x + 1) + 9 Combine like terms. Distributive property = 4x 2 – 8x + 4 + 9 = 4x 2 – 8x + 13 Combine like terms.

  19. Assignment p. 15, 3-39 every 3rd problem (3,6,9,12,…)

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