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3 Functions Problems

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data

    1) Find the equation of the form y = mx+b of the line through the points (2, 3) and the
    midpoint of the line segment joining (−1, 4) and (3, 2).
    2) Calculate the coordinates of the point(s) where the line through the point (−1, 2) with
    slope 1/2 intersects the circle with centre the origin and radius 5.
    3) Find all values of k so that 4kx^2 + 3x + k = 0 has two distinct real roots.

    2. Relevant equations

    1) y=mx+b
    2) I really do not know
    3) Factor and then solve for k

    3. The attempt at a solution

    1) m=(2-4)/(3--1) That is my slope, but what is the y intercept?
    2) I don't really understand the question.
    3) How am I to factor a polynomial with 2 unknowns?


    Please help. I'm really lost.
     
  2. jcsd
  3. Sep 12, 2009 #2
    1) You need to find the coordinate of the midpoint between (−1, 4) and (3, 2). Once you find that, you'll have two points with which to find the equation of the line y = mx + b.

    2) Have you found the equation of the line given? What will be the equation of the circle for the one described in the problem? Find those two equations then solve them simultaneously, or together.

    3) Do you remember something to do with quadratic equations that tells you how many real roots a quadratic has?
     
  4. Sep 12, 2009 #3
    I got 1, thank you.

    For 2, the equation would be 1/2x + 2, correct? I'm stuck at the circle portion, however.

    3) For a quadratic equation, there are 2 roots, found by factoring. However, how do I factor a function with 2 unknowns (k and x)?
     
  5. Sep 12, 2009 #4
    2) You almost have the right equation, but one of the numbers is wrong.
    For a circle with its center at the origin and with radius r, its equation is x2 + y2 = r2

    3) Do you know about the quadratic formula and the discriminant?
     
  6. Sep 12, 2009 #5
    2) Which number is wrong? The slope is given and I think the y intercept is the y-value. as for the circle, would the y-value be the y intercept and would the r be 5? then I can solve for x.

    3) Do I factor out the k and use x=-b+-sqrtb^2-4ac/2a?
     
  7. Sep 12, 2009 #6
    2) Plug in (-1, 2) in your equation and you'll see it's wrong. The slope is correct, so the y-intercept is wrong. Check your math again with solving for b in y = mx + b using (-1, 2).
    Your circle equation will just be x2 + y2 = 52 (since you know r = 5). x and y intercepts aren't important with the circle.

    3) Starting with 4kx^2 + 3x + k = 0, and using the quadratic formula (don't factor out any k), what will a, b, and c be?
     
  8. Sep 12, 2009 #7
    2) I'm getting the y-intercept of 3/2. What do I do with the 5 radius, however?

    3) a will be 4k, b will be 3 and c will be k.
     
  9. Sep 12, 2009 #8
    2) 3/2 is still wrong. Try showing your work in finding b.
    Since you know the radius is 5, you plug it into the circle equation. It's in my previous post. You solve that equation with the correct line equation once you've got it.

    3) Now plug the three values into the discriminant; are you familiar with the discriminant?
     
  10. Sep 12, 2009 #9
    2) 2=1/2(-1)+b ? makes sense?

    3) Not familiar I'm afraid.

    Edit: since b2-4ac = 1, there are two roots. Just learned about discriminants. Can you help me from there, please?
     
    Last edited: Sep 12, 2009
  11. Sep 12, 2009 #10
    That is right so far. but I suspect that when you add 1/2 and 2, you're rewriting 2 as 2/2 instead of 4/2. What should you get?

    3) Plug in the values for a, b, and c in the discriminant √(b2 - 4ac)
    You'll have the unknown k in there, and in order to have two distinct real roots, the discriminant must be greater than 0. How can you find what k should be in order to get two distinct real roots?
     
  12. Sep 12, 2009 #11
    2) 5/2. I subtracted instead of adding...

    3) So k has to be a positive number?
     
  13. Sep 12, 2009 #12
    Can you finish 2) now?

    3) The discriminant has to be greater than 0 to have two distinct real roots. So you want to solve for k in the inequality discriminant > 0 where the discriminant has the values given to you in the original equation. Does that make sense?
     
  14. Sep 12, 2009 #13
    2) What am I to do with the radius though?

    3) I'm not too clear on this. Could you give an example?
     
  15. Sep 13, 2009 #14
    2) The circle's equation is:

    x2-y2=25

    You also got the equation of line:

    y=(1/2)x+n

    to find n substitute for the point (−1, 2)

    When you got both of the equations, you need to solve the system of equations to find the intersection point.

    3)To have two real and distinct roots than the determinant D≰0. In other words D>0.

    You have the equation

    [tex]ax^2+bx+c=0[/tex]

    [tex]D=b^2-4ac >0[/tex]

    Just find a,b,c in your equation. :smile:
     
  16. Sep 13, 2009 #15
    2) How do I go about solving the system of equations?

    3) How can I find a,b,c when I don't know what K is?
     
  17. Sep 13, 2009 #16
    2) In a case like this, the best way to solve the system of equations is to use substitution. You have the equation y = (1/2)x + 5/2 and you know what y is since it's by itself. Substitute, or replace y in the circle equation x2 + y2 = 25 with what it is equal to, so you have an equation with just one variable: x. Then you can solve for that variable. Do that first and post what you get.

    3) The discriminant is b2 - 4ac (the square root isn't important here) and earlier you said "a will be 4k, b will be 3 and c will be k." Replace those in the discriminant in the inequality b2 - 4ac > 0 and post what you get. Do you see why you would want to get something like this for the problem?
     
  18. Sep 14, 2009 #17
    3) So it will be (3)^2-4(4k)(k) which is 9-16k^2 > 0.
     
  19. Sep 15, 2009 #18
    Yes, so now you have to solve for k, and you will have the values for k that give the equation two distinct real roots.
     
  20. Sep 15, 2009 #19

    Mark44

    Staff: Mentor

    This is not the equation of any circle.
     
  21. Sep 15, 2009 #20
    You could let him spot that I wrote the wrong sign. Sigh.
     
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