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    Question: The curve y=f(x) passes through the origin.
    Given that \frac{dy}{dx} = 3x^{2} - 8x -5, find the coordinates of the other points where the curve crosses the x-axis.

    So i started integrating to get y=x^{3}-4x^{2}- 5x+c. Since the curve goes through origin, i assume that c=0. how do i find the co-ordinates of the other points where the curve crosses the x-axis.

    Should i just just substitute random number like 1 or 2 for x and work out what their y co-ordinate is?
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    (Original post by Chelsea12345)
    Question: The curve y=f(x) passes through the origin.
    Given that \frac{dy}{dx} = 3x^{2} - 8x -5, find the coordinates of the other points where the curve crosses the x-axis.

    So i started integrating to get y=x^{3}-4x^{2}- 5x+c. Since the curve goes through origin, i assume that c=0. how do i find the co-ordinates of the other points where the curve crosses the x-axis.

    Should i just just substitute random number like 1 or 2 for x and work out what their y co-ordinate is?
    You're trying to solve x^3-4x^2-5x=0. Try factorizing that. There is actually no need to use the factor theorem.

    hint
    If you were to have x^5-25x=0 you could say that x(x^4-25)=0 \Rightarrow x=0, \pm \sqrt{5}.
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    x(x-5)(x+1)
    So it crosses at (0,0) (5,0) (-1,0)
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    (Original post by Kvothe the Arcane)
    You're trying to solve x^3-4x^2-5x=0. Try factorizing that. There is actually no need to use the factor theorem.

    hint
    If you were to have x^5-25x=0 you could say that x(x^4-25)=0 \Rightarrow x=0, \pm \sqrt{5}.
    Thankyou for the hint btw! so i took out the x to x(x^2 -4x -5)=0.
    Then i added 5 to the other side to get x^2 - 4x = 5. What should i do now to solve this?
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    (Original post by Chelsea12345)
    Thankyou for the hint btw! so i took out the x to x(x^2 -4x -5)=0.
    Then i added 5 to the other side to get x^2 - 4x = 5. What should i do now to solve this?
    You are welcome.
    Do you know how you'd factorize the quadratic equation x^2-4x-5=0?
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    (Original post by Kvothe the Arcane)
    You are welcome.
    Do you know how you'd factorize the quadratic equation x^2-4x-5=0?
    Yes i didnt realise it was a quadratic inside the brackets!
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    (Original post by Mr Moon Man)
    x(x-5)(x+1)
    So it crosses at (0,0) (5,0) (-1,0)
    Not cool. Please read the Guide To Posting at the top of the Maths Forum and pay close attention to the bit about not providing full solutions.
 
 
 
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