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Math Help - Please check over my basic algebra p.I

  1. #1
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    Post Please check over my basic algebra p.I

    Hello there first time poster. I'm currently on my second lesson of math tuition and have been set some questions. I'm not here to cheat I have had a crack at solving them. I would appreciate if people could confirm whether the answers are right and if there's a better method of getting the answer. Thankyou

    Question 1. Now Complete
    x^2  +  2x  - 63
    Factorise
    x^2  = (0+7)  (0-9)
    x = -7
    x =  9 [these are the x crossing points]

    Y crossing point
    y= x^2 + 2x - 63
    0^2 + 2.0 - 63
    y = -63

    minimum point of x (Diff)
    x^2 + 2x
    x^1.2 + 2x^1.1
    2x + 2x
    2x -2 = 0
    -2 = 2x
    x = -1

    y's mimum point putting back into equation
    y = x^2 + 2x -63
    -1^2 + 2 x -1 -63
    1+ -2 -63
    y = -60 minumum point
    Uh oh I went wrong above
    It should be: y=x^2=2x-63
    y(-1)=(-1)^2+2.(-1)-63  = 1-2-63=-64[thankyou earboth]
    You calculate actually the y-coordinate of the vertex of the parabola which is indeed the absolute minimum of all possible y-values.
    As you may have noticed the vertex is below the y-intercept (0, -60).
    Sketch:



    Question 2.currently working on
    x^2-8x+16

    Factorise
    X^2= (0+4) (0-4)
    x = -4
    x =  4 [these are the x crossing points]

    Y crossing point
     y= 0^2-6 =16
    y=0^2-8.0+16=16
    16 is y

    minimum point of x (Diff)
    x^2-8x
    x^1.2-8x^1.1
    2x-10x
    2x-10=0
    x=-5

    y's mimum point putting back into equation
    X^2-8x+16
     -5-8.-5+16
    [y's minimum point is 51] had assistance of calculator for this bit




    Question 3.not complete
    x^2-3x-4

    Factorise
    X^2= (0+) (0+4)
    x = -1
    x =  -4 [these are the x crossing points...I think]

    Y crossing point
     y= 0^2+5.0+4=4
     y=x^2-3x-4
     02-3.0=-4

    minimum point of x (Diff)
    x^2-3x
    x^1.2-3x^1.1
    2x-3x
    2x-3x=0
    -3=2x
    <b>x=-3/2</b>

    y's mimum point putting back into equation
    X^2-3x-4
    i have changed this into a decimal point, -2.25 - -4.5 = -1.75
    [y's minimum point is -1.75] I believe i have got this wrong because i have not written -3/2^2 or -2.25^2 So -2.25^2 is 5.0625 - -4.5 - 4= 5.5625 (is this y?!)



    I have redone the orginal question with the correction so other users can see where I went wrong and maybe learn from it .


    thankyou for you time , burgoyne
    Last edited by Burgoyne; April 3rd 2008 at 05:23 AM.
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  2. #2
    Senior Member topher0805's Avatar
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    To factor an expression does not mean to find what values of x make the expression equal to 0. Simply recognize that you can use the sum-product method to find the roots and then leave it in factored form:

    (x-9)(x+7)

    Your method and answers for finding the y-intercept are correct.

    Can you please explain what you mean by "minimum point of x"?

    Same thing for "y's mimum point putting back into equation".
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  3. #3
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    Thanks for the reply. Basically I have to create a graph from all my findings.

    The minimum points are for the graph. I found the x minimum point and put it back into the equation as x to find the y minimum point.

    here is a picture (done in paint)
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  4. #4
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    earboth's Avatar
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    Quote Originally Posted by Burgoyne View Post
    Thanks for the reply. Basically I have to create a graph from all my findings.

    The minimum points are for the graph. I found the x minimum point and put it back into the equation as x to find the y minimum point.

    here is a picture
    If you have

    y = x^2+2x-63 and x = -1 (which is correct by the way!) then you get:

    y(-1) = (-1)^2+2 \cdot (-1) -63 = 1-2-63 = -64

    You calculate actually the y-coordinate of the vertex of the parabola which is indeed the absolute minimum of all possible y-values.
    As you may have noticed the vertex is below the y-intercept (0, -60).


    A personal remark: Don't delete your original question. Other members of the forum may benefit from the thread you started.
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  5. #5
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    Question 2.currently working on
    x^2-8x+16

    Factorise
    X^2= (0+4) (0-4)
    x = -4
    x =  4 [these are the x crossing points]

    Y crossing point
     y= 0^2-6 =16
    y=0^2-8.0+16=16
    16 is y

    minimum point of x (Diff)
    x^2-8x
    x^1.2-8x^1.1
    2x-10x
    2x-10=0
    x=-5

    y's mimum point putting back into equation
    X^2-8x+16
     -5-8.-5+16
    [y's minimum point is 51] had assistance of calculator for this bit




    Question 3.not complete
    x^2-3x-4

    Factorise
    X^2= (0+) (0+4)
    x = -1
    x =  -4 [these are the x crossing points...I think]

    Y crossing point
     y= 0^2+5.0+4=4
     y=x^2-3x-4
     02-3.0=-4

    minimum point of x (Diff)
    x^2-3x
    x^1.2-3x^1.1
    2x-3x
    2x-3x=0
    -3=2x
    <b>x=-3/2</b>

    y's mimum point putting back into equation
    X^2-3x-4
    i have changed this into a decimal point, -2.25 - -4.5 = -1.75
    [y's minimum point is -1.75] I believe i have got this wrong because i have not written -3/2^2 or -2.25^2 So -2.25^2 is 5.0625 - -4.5 - 4= 5.5625 (is this y?!)
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