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Thread: A line with equation x = p

  1. #1
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    A line with equation x = p

    Given is the function fp(x) = 4/x + p With p element of R


    Asked:
    The line with equation x = p intersects the graph of fp(x) in point Sp. Determine the values of the y-coordinate of Sp can take on.


    I come to the next, but that is not good, according to the answer sheet , that gives: Bsp = (<=,-4] and [4,=>)

    I substitute for x p => 4/p + p =0 => 4 + p^2=0 => p= +-4


    Who can see what goes wrong with my development?
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  2. #2
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    Re: A line with equation x = p

    What you have written is y= (4/x)+ p. Sometimes that is, incorrectly, written when a person really means y= 4/(x+ p). Can you verify which you mean?


    Why are you setting y equal to 0? The problem asks for possible values of y, not p.

    If y= (4/x)+ p and x= p then y= (4/p)+ p= (4+p^2)/p. If we attempt to solve that for p, we get 4+ p^2= py or p^2- py+ 4= 0. The "discriminant" from the quadratic formula is p^2- 16. p will be real if and only if p^2- 16\ge 0 or p^2\ge 16. That gives p\le -4 or p\ge 4.
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  3. #3
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    Re: A line with equation x = p

    Quote Originally Posted by HallsofIvy View Post
    What you have written is y= (4/x)+ p. Sometimes that is, incorrectly, written when a person really means y= 4/(x+ p). Can you verify which you mean?


    Why are you setting y equal to 0? The problem asks for possible values of y, not p.

    If y= (4/x)+ p and x= p then y= (4/p)+ p= (4+p^2)/p. If we attempt to solve that for p, we get 4+ p^2= py or p^2- py+ 4= 0. The "discriminant" from the quadratic formula is p^2- 16. p will be real if and only if p^2- 16\ge 0 or p^2\ge 16. That gives p\le -4 or p\ge 4.


    As you wrote the function is correct. I should have written (4/x)

    I do not know how you come at p^2- 16. Can you further explain in detail? Or should I just know it?
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  4. #4
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    Re: A line with equation x = p

    The "quadratic formula", x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a} gives the roots of the quadratic equation ax^2+ bx+ c= 0. x will be real as long as the "discriminant", b^ 2- 4ac, is not negative. That is, b^2- 4ac\ge 0.

    Here, the equation for p is the quadratic equation p^2- yp+ 4= 0 so that, in the formula above, a= 1, b= -y and c= 4. The discriminant, that determines whether there are any real roots is (-y)^2- 4(1)(4)= y^2- 16. The condition that there be real roots is that y^2- 16= (x- 4)(x+ 4) be non-negative. In order that the product of two numbers be non-negative, they either be both non-negative themselves or both negative.
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  5. #5
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    Re: A line with equation x = p

    Hey HallsofIvy


    thanks for your explanation! I get it now.
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