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Math Help - Afew Questions

  1. #1
    Raj
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    Afew Questions

    Any help with the following questions is appreciated

    Linear/Non-Linear Systems of Equations

    1. Write an equation to form a system with y=5x-1 so that the system has (2,9) as its only solution.

    Answer is y=2x+5
    But how do i get that answer?

    2. The range of the graph y=\frac{1}{2}(x-12)^2+15 is: ?



    Quadratic Functions

    1. What is the new equation when y=(x-2)^2+9 is translated 5 units left and 3 units down?


    Quadratic Equations

    1. Determine the values of K so that the graph f(x)=x^2+4x+k crosses the x-axis twice.

    b^2-4ac>0
    (4)^2-4(1)(k)>0
    16-4k>0
    -4k>-16
    k<4

    Im wondering how to tell if i need a greater/less or equal sign (I think it is > for 2 real roots, < 2 unreal, = roots that are equal) also why it switched on the last step from -4k>-16 to K<4

    Polynomial Functions/Equations

    1. What are the zeroes of 7(2x+3)^3(x-1) > 0?

    I got x=\frac{-3}{2},1
    But i think there are others?
    Last edited by Raj; December 15th 2007 at 05:04 PM.
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    I must ask you what in the world that posting means?
    It seems to be just series of meaningless statements.
    What are you on about?
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  3. #3
    Raj
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    Was in a rush sort of, i can't believe i forgot my notes at school >.<

    I think its more clearer now.
    Last edited by Raj; December 15th 2007 at 04:09 PM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Raj View Post
    1. Write an equation to form a system with y=5x-1 so that the system has (2,9) as its only solution.

    Answer is y=2x+5
    But how do i get that answer?
    I presume that you are looking for another linear function, else there are an infinite number of answers.

    So let's assume we have a new equation of the form y = mx + b.

    So solve the system
    y = 2x + 5
    y = mx + b
    such that the solution is (2, 9).

    I'll start you off:
    Put the solution point into the second equation:
    9 = 2m + b

    So
    b = 9 - 2m

    Thus the second equation is
    y = mx + (9 - 2m)

    Now we need to solve the system
    y = 2x + 5
    y = mx + (9 - 2m)
    for m.

    See what you can do with this.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Raj View Post
    2. The range of the graph y=\frac{1}{2}(x-12)^2+15 is: ?
    The range of a graph is the set of all possible y values that the function takes. So what y values can this function take?

    (Hint: Graph it.)

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Raj View Post
    1. What is the new equation when y=(x-2)^2+9 is translated 5 units left and 3 units down?
    Consider the function y = f(x).
    When we translate this graph h units to the right the corresponding function is y = f(x + h).

    When we translate this graph h units to the left the corresponding function is y = f(x - h).

    When we translate this graph k units up the corresponding function is y - k = f(x) \implies y = f(x) + k.

    When we translate this graph k units down the corresponding function is y + k = f(x) \implies y = f(x) - k.

    See if you can apply these rules to get your answer.

    -Dan
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Raj View Post
    1. Determine the values of K so that the graph f(x)=x^2+4x+k crosses the x-axis twice.

    b^2-4ac>0
    (4)^2-4(1)(k)>0
    16-4k>0
    -4k>-16
    k<4

    Im wondering how to tell if i need a greater/less or equal sign (I think it is > for 2 real roots, < 2 unreal, = roots that are equal) also why it switched on the last step from -4k>-16 to K<4
    Note that if the discriminant is equal to 0 then there is only one solution to the quadratic equation. Thus we can't include the equal sign on the line b^2-4ac>0.

    Whenever you divide both sides of an inequality by a negative number the inequality changes "direction." So
    -4k>-16

    \frac{-4k}{-4} < \frac{-16}{-4}

    k<4

    If you have troubles seeing this, consider a simpler example. Solve
    -x > 1

    You can easily get the solution set from the given inequality, and you should be able to see that it is x < -1, which is what you get when you divide both sides by -1.

    -Dan
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Raj View Post
    1. What are the zeroes of 7(2x+3)^3(x-1) > 0?

    I got x=\frac{-3}{2},1
    But i think there are others?
    I presume this is really 7(2x+3)^3(x-1) = 0?

    Why should there be other zeros?

    The concept behind solving something like this is the following theorem:
    If ab = 0 then either a = 0 or b = 0 or both.

    So when we a simpler example like
    (x - 1)(x - 2) = 0
    then either
    x - 1 = 0 \implies x = 1
    or
    x - 2 = 0 \implies x = 2
    (and we can't have both terms equal to 0 at the same time.) You've likely seen this process any number of times when solving a quadratic that factors.

    In this case we have:
    7(2x+3)^3(x-1) > 0

    7(2x + 3)(2x + 3)(2x + 3)(x - 1) = 0

    So either
    7 = 0 <-- Not true!
    or
    2x + 3 = 0 \implies x = -\frac{3}{2}
    or
    2x + 3 = 0 \implies x = -\frac{3}{2}
    or
    2x + 3 = 0 \implies x = -\frac{3}{2}
    or
    x - 1 = 0 \implies x = 1

    So we have
    x = -\frac{3}{2}, -\frac{3}{2}, -\frac{3}{2}, 1

    or more simply
    x = -\frac{3}{2}, 1

    -Dan
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