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Math Help - [SOLVED] Help: Algebra Questions

  1. #1
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    Question [SOLVED] Help: Algebra Questions

    I have a couple of algebra questions which I'm stuck on. Any help would be appreciated.

    #1:

    x/2 + y/7 = 1
    How can I re-arrange it so that is in the form y=

    #2:

    How do you find the turning point of a quadratic equation such as: y=x^2 + 6x + 8

    #3:

    Find the value(s) for k for which 2(x-1)^2 + 2k = 10 has
    a) exactly one real solution
    b) no real solutions

    I don't even know where to start on this one

    ---

    Thanks

    ---

    Thanks for all the help Now I can finally finish my homework.
    Last edited by starship; March 2nd 2008 at 07:59 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by starship View Post
    x/2 + y/7 = 1
    How can I re-arrange it so that is in the form y=
    \frac{x}{2} + \frac{y}{7} = 1

    \frac{y}{7} = -\frac{x}{2} + 1

    Multiply both sides by 7:
    y = -\frac{7x}{2} + 7

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by starship View Post
    #2:

    How do you find the turning point of a quadratic equation such as: y=x^2 + 6x + 8
    You need to put the quadratic equation in vertex form by completing the square:

    y = x^2 + 6x + 8

    y = (x^2 + 6x) + 8

    y = (x^2 + 6x + 9 - 9) + 8

    y = (x^2 + 6x + 9) - 9 + 8

    y = (x + 3)^2 - 1

    The vertex form is
    y = a(x - h)^2 + k
    where the vertex is (h, k), so the vertex of this parabola is at (-3, -1).

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by starship View Post
    #3:

    Find the value(s) for k for which 2(x-1)^2 + 2k = 10 has
    a) exactly one real solution
    b) no real solutions

    I don't even know where to start on this one
    It's probably easiest to expand this:
    2(x - 1)^2 + 2k = 10

    2(x^2 - 2x + 1) + 2k - 10 = 0

    2x^2 - 4x + 2 - 2k - 10 = 0

    2x^2 - 4x + (-8 - 2k) = 0

    Now, the discrimiant of a quadratic tells you what kind of solutions you have.
    D = b^2 - 4ac
    (You might recall this as the quantity under the square root in the quadratic formula.)

    When D > 0 we have two real, unequal solutions
    When D = 0 we have two real equal solutions, or one real solution depending on how you want to look at it.
    When D < 0 we have two complex solutions.

    So figure out the determinant in terms of k and see what happens.

    -Dan
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  5. #5
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    Alternately, for #2: The "turning point" is the vertex. The vertex is always located on the axis of symmetry, which is the line of symmetry, the line which divides the parabola into two identical halves.

    The equation for the line of symmetry is:

    x=-\frac{b}{2a}

    so yours is:

    x=-\frac{6}{2} = -3

    Since the vertex always lies on the line of symmetry, it turns out that your vertex has an x-coordinate of -\frac{b}{2a}.

    To find the y-coordinate, plug the x-coordinate into your equation.

    In your equation:

    y=-3^2+6(-3)+8 = 9 - 18 + 8 = -1.

    Thus, the coordinate for your vertex, as previously noted, is (-3, -1).
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