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Math Help - rearranging equations

  1. #1
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    rearranging equations

    I have no idea :|

    1a) Express  x[squared]-12x+40 In the form (x-p)[squared] + q

    1b) Hence or otherwise find the least value of x[squared] -12x+40


    in addition can somebody tell me how to get power signs in the above box please?

    Thanks in advance ;-)
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  2. #2
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    Hello Van:

    For the first exercise, complete the square.

    (p, q) are the coordinates of the vertex, by the way.

    For the second exercise, I regret that I do not understand [tex]x[squared]. Is this some sort of multiplication?

    Cheers,

    ~ Mark
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  3. #3
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    Ah Yes sorry Mark, i did the tags wrong!!


    (i may be back here :S )
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  4. #4
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    Quote Originally Posted by vanpopta View Post
    I have no idea :|

    1a) Express  x^2-12x+40 In the form (x-p)^ + q

    1b) Hence or otherwise find the least value of x^2 -12x+40


    in addition can somebody tell me how to get power signs in the above box please?

    Thanks in advance ;-)
    Just click on the expression and it will show you the code used to make it.

    1a) I think you just need to 'complete the square'. Are you familiar with this technique?

     x^2-12x+40

    (x^2-12x+36)+40-36

    (x-6)^2+4

    1b) Are you trying to find the minimum value of the parabola?

    y=x^2-12x+40

    Use x=\frac{-b}{2a} using the standard form of the parabola: f(x)=ax^2+bx+c

    This will give you the x-coordinate of the vertex. You want to find the y-coordinate of the vertex, so substitute the x-coordinate into your original equation to solve for y.

    This will be a minimum value since the parabola opens upward.

    x=\frac{-(-12)}{2(1)}=\frac{12}{2}=6

    Now find f(6) in f(x)=x^2-12x+40
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  5. #5
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    Quote Originally Posted by masters View Post

    1a) I think you just need to 'complete the square'. Are you familiar with this technique?

     x^2-12x+40

    (x^2-12x+36)+40-36

    (x-6)^2+4
    I am very bad at this , and am not sure how you got from the first to the second line :S
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  6. #6
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    Quote Originally Posted by masters View Post
    ... Use x=\frac{-b}{2a} using the standard form of the parabola: f(x)=ax^2+bx+c
    Quote Originally Posted by masters View Post

    This will give you the x-coordinate of the vertex ...

    Yes, indeed.

    And, if you would like to use formulas instead of the (p, q) values, then the y-coordinate of the vertex is:

    c - \frac{b^2}{4a}

    Cheers,

    ~ Mark
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  7. #7
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    Hi Van:

    If you do a Google search on keywords "complete the square", then you will find hundreds of lessons and examples.

    Cheers,

    ~ Mark
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  8. #8
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    Quote Originally Posted by mmm4444bot View Post


    Yes, indeed.

    And, if you would like to use formulas instead of the (p, q) values, then the y-coordinate of the vertex is:

    c - \frac{b^2}{4a}

    Cheers,

    ~ Mark
    Lost my head. Did more work than I needed to. Forgot that I already completed the square on this one. It's still a good thing to know.
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  9. #9
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    Thanks very much so far guys, i'm stuck again though


    2(b) Sketch the Graph of y=x^2+10x+20 Stating the co-ordinates of the vertex.

    I know its either a U or n shape, and the vertex is the extreme point on the curve, but i'm not sure how exactly to do it.


    3(b) Find the constants a, b and c such that for all values of x

     3x^2-6x+10 = a(x-b)^2+c

    I know i asked a similar question already, but i'm stumped
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  10. #10
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    Complete the square, complete the square, complete the square ...



    Did you understand my first post when I told you that (p, q) are the coordinates of the vertex?

    In other words, after you complete the square, you can immediately plot the vertex point.

    You can see the y-intercept by simply inspecting the original equation, so you can immediately plot that point without doing any algebra at all.

    After completing the square on this exercise, the two x-intercepts are found quite easily by setting y equal to zero and solving for x. (i.e., there's no need to use the quadratic formula.)

    Plot those two points, and you've got four points plotted.

    Sketch the right half of the parabola from the vertex through the rightmost x-intercept and up through the y-intercept.

    Use symmetry, along with the other x-intercept, to sketch the left half.

    You're done!

    (By the way, when the leading coefficient is positive in a quadratic equation, then the parabola opens upward, and when the leading coefficient is negative, then the parabola opens downward.)

    On your second exercise, do you recognize that the right side is the form that you get after completing the square?

    In other words, complete the square on that exercise also, and you can then immediately read off the values of a, b, and c.

    NOTE: Whoever authored that exercise should have stuck with the symbols p and q (or anything other than b and c) because the symbols b and c are NOT the same as the coefficients in the familiar ax^2 + bx + c form. Instead, the b and c are the coordinates of the vertex: (b,c). I think using b and c is quite potentially confusing. The symbol a is actually the leading coefficient, though.

    Let us know if you need help completing the square on the second exercise. Remember to factor out the leading coefficient before you start. (Did you check out any lessons or examples?)

    Cheers,

    ~ Mark
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