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Homework Help: Intersection of two graphs

  1. Nov 21, 2009 #1
    the graph of y=kx+3 intersects the graph of y=x^2 +8x at two distinct points for
    'k 'equals what?

    to be honest I do not know where to start
     
  2. jcsd
  3. Nov 21, 2009 #2
    It's a parabola and a line.
     
  4. Nov 21, 2009 #3
    yeah....but how do you find k
     
  5. Nov 21, 2009 #4

    Mark44

    Staff: Mentor

    At any point of intersection of the two curves, the coordinates x and y will satisfy both equations.

    You have y=kx+3 and y = x^2 + 8x. Set the right sides of these equations equal and solve for x.
     
  6. Nov 21, 2009 #5
    how can you solve for x when you dont know k
     
  7. Nov 21, 2009 #6
    You're not finding an exact value for x, just getting it by itself and leaving the equation in terms of k.
     
  8. Nov 21, 2009 #7

    Mentallic

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    x will be in terms of k.
    Once you set the quadratic in the form [itex]ax^2+bx+c[/itex] with a,b,c any number or variable in terms of k, use the quadratic formula to solve for x.

    Now, once you have x=... what do you need in order to have 2 distinct points? It is the same thing as thinking of what you would need to have 2 distinct roots to a quadratic equation.
     
  9. Nov 21, 2009 #8
    Can anyone actually show me their working out so I can gain a more coherent understanding?
     
  10. Nov 22, 2009 #9

    HallsofIvy

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    Better if you show us your working. You learn math by doing, not by watching!

    As Mark44 told you to start with, set y= kx+ 3= x^2+ 8x. Write that in the form ax^2+ bx+ c= 0 and use the quadratic formula. Yes, there will be a k in the formula.

    You don't have to actually solve the equation to answer this question. Since the question asks about two different points of intersection, you want to decide what k must be so that equation has two distince solutions. A quadratic equation can have 0, 1, or 2 solutions depending on the "discriminant", b^2- 4ac.
     
  11. Nov 22, 2009 #10
  12. Nov 22, 2009 #11
    so once rearranged it will be x^2+(8x-kx)-3=0
    Use the quadratic formula and it is: (-8+k +/- sqrt k^2-16k+76 ) / 2
    am i correct up to that point?
     
  13. Nov 22, 2009 #12

    HallsofIvy

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    Yes, what you have is correct.

    And, as I said, you really only need the discriminant: (8- k)^2+ 12= k^2-16k+ 76. For what values of k is that positive?
     
  14. Nov 22, 2009 #13

    Mark44

    Staff: Mentor

    You are solving for x, so you really should have x on one side of an equation.

    x = [(-8 + k) +/- sqrt(k^2 - 16k + 76)]/2
     
  15. Nov 22, 2009 #14
    Sorry HallsofIvy I am having trouble understanding what you meant by "For what values of k is that positive".

    the discriminant is k^2-16k+76, what do you do with that
     
  16. Nov 22, 2009 #15

    Mark44

    Staff: Mentor

    You have k^2 - 16k + 76 as the discriminant. For what values of k is the discriminant > 0? That's what HallsOfIvy asked. Do you know how to solve a quadratic inequality?
     
  17. Nov 22, 2009 #16
    umm..unfortunately i don't
     
  18. Nov 22, 2009 #17

    Mark44

    Staff: Mentor

    Do you know how to solve a quadratic equation?
     
  19. Nov 22, 2009 #18
    By using the quadratic formula?
     
  20. Nov 22, 2009 #19

    Mark44

    Staff: Mentor

    Or however.
     
  21. Nov 22, 2009 #20
    okay so i have k^2-16k+76, if i use the quadratic formula it is (16+/-sqrt (-16)^2 - 4*1*76 )/ 2, but this gives an error on my calculator
     
  22. Nov 22, 2009 #21

    Mentallic

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    Just like when you solve any quadratic, when you get the error (which means the discriminant is negative, and you can't take the square root of a negative), how does this describe the properties of the parabola?

    For e.g. Take a look at the graph of [itex]y=x^2-1[/itex]
    and solve x in [itex]x^2-1=0[/itex]

    Now take a look at [itex]y=x^2+1[/itex]
    and solve x in [itex]x^2+1=0[/itex]

    can you see what is happening and understand why you get an error on the second one?
    When you're solving for x, you're basically saying where does y=0.
     
  23. Nov 22, 2009 #22
    yeah okay i get that, so how am i suppose to get value/s for k
     
  24. Nov 23, 2009 #23

    Mentallic

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    Well, for what values of x is [itex]x^2+1>0[/itex] ?

    In a similar fashion, the discriminant was [itex](k-8)^2+12[/itex]. For what values of k is [itex](k-8)^2+12>0[/itex]. And the answer to this gives you the values of k where the quadratic and the line has 2 distinct roots.
     
  25. Nov 23, 2009 #24
    ohk, so (k-8)^2+12>0
    (k-8)^2>-12

    then what would do since you cannot square root a negative
     
  26. Nov 23, 2009 #25

    Mentallic

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    No, no. I'm just asking you to use logic here.
    What are the possible values of [itex]x^2[/itex] for all x? In other words, what is the range of [itex]x^2[/itex] for all values of x?
    Now, similarly to that, what is the range of [itex](k-8)^2[/itex] for all values of k?

    What if you add 12 to this range now? So finally, for what values of k is [itex](k-8)^2+12>0[/itex]?
     
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