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Homework Help: Algebra problem

  1. Sep 19, 2006 #1
    Algebra problem!!!

    theres this question that i was given in my class notes and she wants us to try and solve it but i just dont get it and i need your help please. I think that you have to use completing the square method but im not sure, but heres the question

    Find the value of K for which the line y=2x+K is a tangent to the curve y=x²-2x-7.

    please help me it would be much appreciated :smile:
  2. jcsd
  3. Sep 19, 2006 #2
    edit: Sorry, what I said before was wrong. Actually, the answer is -11.
    Last edited: Sep 19, 2006
  4. Sep 19, 2006 #3
    okay...in algebra you learn something about stuff like that also.

    try to find 2 points on y=x^2-2x-7 that are equdistant from the vertex of the parabola. the roots of the equation would work. use midpoint formula to find the coordonate of the vertex, and plug it in f(x) to find it's y-coordonate.

    now that you know the y-coordonate of the vertex, that's the K you want.

    y'=2x+k...well the "2x" part is given to you ,meaning that you don't actaully know how to make the x^2 turn into a 2x. Looks like an algebra2/precalc problem so use what i wrote above.
  5. Sep 19, 2006 #4
    Wait a minute... If you do this then you get K=-8 but K should be -11!
    Last edited: Sep 19, 2006
  6. Sep 19, 2006 #5
    OK, I know how to do this. Lets visualize the given curve and the tangent line. Since you already know the slope of the tangent line, all you have to do to get the equation of the tangent line is to "move" your tangent line (without changing its slope) up or down such that it intersects with the curve. BUT, the tagent line should intersect with the curve only once if it is suppose to be called a "tangent."

    Now we know that if curve and the tangent line intersect at a single point then they should have the same y coordinate. So we can set the two equations equal to each other i.e.
    [tex]2x + K = x^2 - 2x - 7[/tex]

    If we solve for x, we get (using the quadratic equation):

    [tex] x = \frac{4 \pm \sqrt{44 + 4K}}{2}[/tex]

    Now since we are looking for a single soltion for x (i.e. only one intersection) then the value inside the sqrt should be zero. The only way this is possible is when K = -11. :smile:
    Last edited: Sep 19, 2006
  7. Sep 19, 2006 #6
    ok i kind of understand but how do you get 44 in the quadratic equation
  8. Sep 19, 2006 #7
    Lets start from the top.

    So you have

    [tex]2x + K = x^2 - 2x - 7[/tex]

    Bringing everything to one side we get:

    [tex]x^2 - 4x - 7 - K = 0[/tex]


    [tex]x^2 - 4x - (7 + K) = 0[/tex]

    This is of the form [itex]ax^2 + bx + c[/itex] where a = 1, b=-4, and c = -(7+K).

    Now you know that we can solve these types of equations using the quadratic forumula which says that

    [tex] x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

    Using the particular values for a, b and c we get:

    [tex] x = \frac{4 \pm \sqrt{16 + 4(7+K)}}{2}[/tex]

    further simplifying we get:

    [tex] x = \frac{4 \pm \sqrt{44+4K}}{2}[/tex]

    Now the only way we can have a single unique solution for x is when [itex]44+ 4K = 0[/itex] which means that K = -11.
    Last edited: Sep 19, 2006
  9. Sep 19, 2006 #8
    thank you for breaking it down to me i finally understand it at last. thank you i really appreciate your help
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