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Warm-Up

Warm-Up. Answers: 1. Vertex (4,-1) 2. Roots are x=3, x= -4 Y-int: (0,-15) vertex: (1, -16). Find the vertex, the roots or the y-intercept of the following forms: 1. f(x) = (x-4) 2 -1 2. f(x) = -2(x-3)(x+4) 3. f(x) = x 2 -2x -15. Classwork: Pg. 309 (3-36 every 3 rd one).

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Warm-Up

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  1. Warm-Up • Answers: • 1. Vertex (4,-1) • 2. Roots are x=3, x= -4 • Y-int: (0,-15) • vertex: (1, -16) Find the vertex, the roots or the y-intercept of the following forms: 1. f(x) = (x-4)2-1 2. f(x) = -2(x-3)(x+4) 3. f(x) = x2-2x -15 Classwork: Pg. 309 (3-36 every 3rd one)

  2. Lesson 3.1A Quadratic Functions– The Three Forms • 1. Standard or General Form y = ax2+ bx + c This form tells me the __________________. That is where the graph ________________. The y-intercept is _____________________. y-intercept crosses the y-axis (0,c)

  3. 2. Factored Form y = a( x– r1) (x-r2) This form tells me the __________________. That is where the graph ________________. To find the roots, zeros or x-intercepts _______________. roots, zeros, or x-intercepts crosses the x-axis Set ( x – r1) (x-r2) = 0

  4. 3. Vertex Form y = a( x–h)2+k This form tells me the __________________. That is where the graph ________________. The vertex is _____________________. vertex Minimum or maximum of the parabola (h,k)

  5. How to convert from different forms? Given:

  6. Given:

  7. Given:

  8. Ex. 1: y = x2-6x + 5 General Form • What form? • Change from general to ____________ by using either _______________ or ___________________. Factored Form Factoring Quadratic Formula X2 – 6x + 5 (x-5)(x-1) = 0 X = 5, x = 1

  9. Ex. 2: y = x2-6x + 5 Vertex form • Change from general form to __________ by using _________________________. Completing the square y = x2-6x + 5 Complete the square Y=(x2-6x + _____) +5 - _____ 9 9 Y=(x-3)2 -4 Now Graph: Y-intercept:_______ Vertex :_______ X-intercepts: ______, ______ (0,5) (3,-4) (5,0) (1,0)

  10. Ex. 3: Using f(x) = -3x2 +6x - 13find the vertex when given the standard form? • Use the formula to find x : • Substitute in x to find y. In standard form a = -3, b = 6, c= -13 x = 1 , then y = -3(1)2 +6(1) -13 = -10 Vertex = (1, -10)

  11. You try: y = (x+4)2-13 • Answers: • GF: x2+8x+3 • FF: y = (x+.39) (x+7.61) HINT: If you cannot factor it, you must use the Quadratic Formula!!

  12. Ex. 4: Minimum and Maximum Quadratic Functions Consider the quadratic function: f(x) = -3x2 + 6x – 13 • Determine, without graphing, whether the function has a minimum value or a maximum value. • Find the max. or min. value and determine where it occurs • Identify the function’s domain and range.

  13. Ex. 4 Continuedf(x) = -3x2 + 6x – 13 Begin by identifying a, b, and c. • Because a < 0, the function has a max. value. If a>0 then the function would have a min. • The max. occurs at x = -b/2a = - 6 /2(-3) = -6/-6 = 1. The maximum value occurs at x = 1 and the maximum value of f(x) = -3x2 + 6x – 13 f(1) = -3*12 + 6*1 – 13 = -3 + 6 – 13=-10 Plug in one for x into original function. We see that the max is -10 at x = 1. c. Domain is (-∞,∞) Range (-∞,-10].

  14. YOU TRY!! Repeat parts a through c using the function: f(x) = 4x2 – 16x + 1000. Answer: • Min • Min is 984 at x = 2 • Domain is (-∞,∞) Range is [984,∞)

  15. Summary: • Describe how to find a parabola’s vertex if its equation is expressed in standard form.

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