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Thread: need some help with quadratic functions

  1. #1
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    need help with THIS quadratic function!

    actually i need help on this quadratic function

    f(x) = -x2 + 10x

    here is my work

    f(x) = -x2 + 10x
    f(x) = -x2 + 10x + 25 - 25 ((i divided 10 with 2 and then i squared it to get 25, then i added and subtracted 25 to keep the function even))
    f(x) = -1(x - 5)2 - 25

    here is where i am stuck... how can i factor -x2 + 10x + 25?
    Last edited by abel muroi; 01-28-2015 at 09:43 PM.

  2. #2
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by abel muroi View Post
    actually i need help on this quadratic function

    f(x) = -x2 + 10x
    What were the instructions for this? What are you supposed to be doing with this? Thank you!

  3. #3
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    Quote Originally Posted by abel muroi View Post
    actually i need help on this quadratic function

    f(x) = -x2 + 10x

    here is my work

    f(x) = -x2 + 10x
    f(x) = -x2 + 10x + 25 - 25 ((i divided 10 with 2 and then i squared it to get 25, then i added and subtracted 25 to keep the function even))
    f(x) = -1(x - 5)2 - 25 should be + 25 not -25

    here is where i am stuck... how can i factor -x2 + 10x + 25?
    I guess (but should NOT have to) that you want to put this into standard form (at least that is what I think it is called).
    OK we will assume that you did not make any mistakes for a moment. YOU DO NOT KNOW WHAT AN EQUAL SIGN MEANS! This is not good. You ask how you can factor -x2 + 10x + 25 which YOU claim equals -x2 + 10x. So just factor -x2 + 10x! After all it equals -1(x - 5)2 - 25

    Now onto your mistake. As I told you in the past you only add on the 1/2 of the the coefficient of x squared only if the coefficient of x^2 is 1. You have -1.

    Try this: -x^2 + 10x = -1(x^2 -10x) . Now determine what you need to do to make a perfect square.

    Why do you want to factor the end result anyways? I have a better question for you. What is the vertex of this parabola (quadratic)?
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  4. #4
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    Quote Originally Posted by Jomo View Post
    I guess (but should NOT have to) that you want to put this into standard form (at least that is what I think it is called).
    OK we will assume that you did not make any mistakes for a moment. YOU DO NOT KNOW WHAT AN EQUAL SIGN MEANS! This is not good. You ask how you can factor -x2 + 10x + 25 which YOU claim equals -x2 + 10x. So just factor -x2 + 10x! After all it equals -1(x - 5)2 - 25

    Now onto your mistake. As I told you in the past you only add on the 1/2 of the the coefficient of x squared only if the coefficient of x^2 is 1. You have -1.

    Try this: -x^2 + 10x = -1(x^2 -10x) . Now determine what you need to do to make a perfect square.

    Why do you want to factor the end result anyways? I have a better question for you. What is the vertex of this parabola (quadratic)?
    -1(x2 + 10x) = -1(x + 5)2 + 25

    did i do this right?

    so the vertex should be -5, 25?

  5. #5
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    Quote Originally Posted by abel muroi View Post
    -1(x2 + 10x) = -1(x + 5)2 + 25

    did i do this right?

    so the vertex should be -5, 25?
    Everything you wrote is correct but does not match the original equation that you gave us. Also -5,25 is not the vertex which is a point. (-5,25) is the vertex.


    -x2 + 10x does NOT equal -1(x2 + 10x) as this equals -x2 - 10x

    Please try again
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  6. #6
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    Quote Originally Posted by abel muroi View Post
    f(x) = -x2 + 10x
    here is my work

    f(x) = -x2 + 10x
    f(x) = -x2 + 10x + 25 - 25
    (i divided 10 with 2 and then i squared it to get 25, then i added and subtracted 25 to keep the function even)
    f(x) = -1(x - 5)2 - 25

    here is where i am stuck... how can i factor -x2 + 10x + 25?
    f(x) = -x2 + 10x

    Abel, before running around like a chicken with its head cut off,
    try getting started this way:
    -x^2 + 10x = 0
    x(-x + 10) = 0
    x = 0
    or
    -x + 10 = 0
    x = 10

    Now at least you sort of know where you're standing...

    Do you have graph paper?
    If no, BUY SOME; if yes, USE IT
    I'm just an imagination of your figment !

  7. #7
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    Quote Originally Posted by Jomo View Post
    Everything you wrote is correct but does not match the original equation that you gave us. Also -5,25 is not the vertex which is a point. (-5,25) is the vertex.


    -x2 + 10x does NOT equal -1(x2 + 10x) as this equals -x2 - 10x

    Please try again
    what about -1(x2 - 10x)?

    also, if the quadratic is in the form -1(x2 + 10x).. do i have to keep factoring until i have the standard form right?

    and with the standard form i can get the vertex of the parabola.

  8. #8
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    Quote Originally Posted by abel muroi View Post
    what about -1(x2 - 10x)?

    also, if the quadratic is in the form -1(x2 + 10x).. do i have to keep factoring until i have the standard form right?

    and with the standard form i can get the vertex of the parabola.
    You want your answer in the form f(x) =a(x +/- b)^2 + c. Your a=-1.

    f(x) = -x2 + 10x = -(x^2 - 10x)
    complete the square inside the parenthesis. [(1/2)(-10)]^2 = [-5]^2 = +25
    f(x) = -(x^2 - 10x + 25) +25 (the +25 outside the parenthesis cancelled out the +25 I wrote inside the parenthesis. Can you see why.
    So f(x) = -( x - 5)^2 +25. So the vertex is (5, 25) and the y-intercept is 0
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  9. #9
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    Quote Originally Posted by abel muroi View Post
    actually i need help on this quadratic function

    f(x) = -x2 + 10x

    here is my work

    f(x) = -x2 + 10x
    f(x) = -x2 + 10x + 25 - 25 ((i divided 10 with 2 and then i squared it to get 25, then i added and subtracted 25 to keep the function even))
    f(x) = -1(x - 5)2 - 25

    here is where i am stuck... how can i factor -x2 + 10x + 25?
    Hi abel,

    It would generally help if we knew what you were trying to do. For example, as suggested by Jomo, suppose you were trying to put the equation into a standard form so you could determine the vertex. If so, and for this equation in particular, there is another way which might be faster on a test and thus, be worth knowing.

    For example, the x value of the vertex is midway between the zero's of the equation. You can show this with the solutions from the quadratic formula. You also get this from completing the square when you divide the x coefficient by two. That value is the negative of the x value of the vertex when the coefficient of the x2 term is 1. So, in general, we have for any quadratic
    f(x) = a x2 + b x + c = a (x2 + (b/a) x + c/a) )
    and we have the vertex is at x = -b/(2a)

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