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Grade 11 Quadratics help

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  1. May 11, 2010 #1
    1. The problem statement, all variables and given/known data
    f(x) : ax²+bx+c, a≠ 0

    [have to find the a(x-h)²+k form first]

    Find an expression in terms of a, b, c for: (without using graph or calculator)
    (i) An equation of the axis of symmetry
    (ii) The maximum or minimum value
    (iii) The coordinated of the vertex
    (iv) The domain and the range
    (v) The y-intercept of the graph of the function
    (vi) The zeroes of the function

    Discuss how you can predict the number of zeroes for a given quadratic function of the form y= ax²+bx+c, a≠ 0. Support the validit of your prediction with some examples.



    My attempt was
    (ax² + bx + b/2 - b/2) + c
    (ax² + bx + b/2) + c - b/2
    and after this i'm stuck

    (v) y intercept is (0,c)
     
  2. jcsd
  3. May 11, 2010 #2

    rock.freak667

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    Homework Helper

    ax2+bx+c=A(x-h)2+k and expand the right side and then equate coefficients if you're unable to do it the other way.

    Try adding and subtracting b/4a2 instead of b/2.
     
  4. May 11, 2010 #3
    umm thanks:) but i cant make ax2+bx+c=A(x-h)2+k because ax2+bx+c is like an equation just like 3x2-6x+16
    and can you please explain me how did you come up with b/4a2
     
  5. May 11, 2010 #4

    rock.freak667

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    Homework Helper

    I meant expand out A(x-h)2 and then equate 'a' to whatever the coefficient of x2 is in the expansion.

    Sorry, I was supposed to type b2/4a2. Well remember that you want to have a perfect square (x-h)2 which give x2-2hx+h2 when expanded out.
     
  6. May 11, 2010 #5

    Mark44

    Staff: Mentor

    No, ax2+bx+c is not "like" an equation; an equation can be recognized by the presence of the = sign. ax2+bx+c is an expression, just as 3x2-6x+16 is.

    Since you are NOT working with an equation, you are very limited in the things you can do. You can't add the same amount to both sides, because there are not two sides. About all you can do is add 0.

    You want to complete the square by adding a certain amount and subtracting exactly the same amount so that the net change is 0.

    For example,
    22 + 4x
    = 2(x2 + 2x)
    = 2(x2 + 2x + 1) - 2
    = 2(x + 1)2 - 2

    In the 3rd line I added 1 inside the parentheses, but I really added 2, so to keep the expression equal to the previous one, I had to balance things by adding -2 (or subtracting 2).
     
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