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How do u figure this out?

  1. Dec 24, 2004 #1
    If f(x)=x^(2)-2x-8 then what is the y-intercept of y=-f(x)+2?

    I dont know how to figure this out. I have the answer but dont know how to get it.
     
  2. jcsd
  3. Dec 24, 2004 #2

    quasar987

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    I'm assuming the "y-intercept" is referring ot the y-coordinates at which the two curves intersect each other?

    In which case, it is always the same method to find the points of intersection between two functions:

    1) you set them equal with one another

    2) solve for x

    This gives you the "x intersects". To find the y intercept, just plug in the x intercept in either of the two functions (since they are equal at those precise points) and this gives you the corresponding y-intercept.
     
  4. Dec 24, 2004 #3

    quasar987

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    Now I'm thinking maybe the "y-intercept" refers to the points at which the curve intercept the y axis...

    In this case, ok... f(x) = x^(2)-2x-8 and y = -f(x)+2. That is to say, y = -(x^(2)-2x-8)+2 = (-x^(2)+2x+8)+2 = -x^(2)+2x+10.

    When does a function intercept the y axis? It is when x = 0. Just plug x = 0 in your function and it gives you the y coordinate at which the function intercept the y axis.
     
  5. Dec 24, 2004 #4
    Quasars got it right in his second post. The y-intercept is the point at which the curve crossing the y-axis. If it crosses the y-axis, x = 0. I think you can take it from there.
     
  6. Dec 24, 2004 #5
    I still dont get it , I tried doing that but didnt get the right answer the answer is (-2.3,0) and (4.3,0)
     
  7. Dec 24, 2004 #6

    Doc Al

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    solve the quadratic

    Did you solve the quadratic to find where y = 0?
     
  8. Dec 24, 2004 #7
    ok when we factor the quadratic we get x=4 and x=-2
    these are the x intercepts (4,0) and (-2,0) so I flipped them around to get the y intercept (0,4) and (0,-2) now the question says then what is the y intercept of y=-f(x)+2 so I multiplied the y coordinate by -1 and added 2 my answer was (0,-2) and (0,4) is this correct? If so is there an easier way? My answer in the lesson says (-2.3,0) and (4.3,0) is this wrong?
     
  9. Dec 24, 2004 #8

    Doc Al

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    The answer given is correct; your factoring is wrong. (Multiply it out and see.) You'll need to use the quadratic formula.
     
  10. Dec 24, 2004 #9
    Im stuck I dont know what to do where am I using the quadratic formula? With which equation ?
     
  11. Dec 24, 2004 #10

    Doc Al

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    In post #3, quasar987 gave you the expression for y as a function of x. Set y = 0, and solve the resulting quadratic.
     
  12. Dec 24, 2004 #11
    ok so a=-1 b=2 and c=10?

    I got x=-2.16 and x=4.16 the answer said (-2.3,0) and (4.3,0) where did I go wrong? And how come we are solving for x intercepts when the question asked for y-intercepts. :confused:
     
    Last edited: Dec 24, 2004
  13. Dec 24, 2004 #12
    why do we need to solve for the x intercepts? Dont we just need to know the y intercept?

    f(x) = x^2-2x-8

    we need to find y intercept of y = -f(x) + 2

    plug in f(x) and as mentioned above, just plug in 0 for x to solve for y intercept. I don't see why the x intercept is needed here? Or did aisha mistake y intercepts for x intercepts?
     
  14. Dec 24, 2004 #13
    you tell me...
     
  15. Dec 25, 2004 #14

    Doc Al

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    Right.
    Be more careful with your arithmetic!
    Good point! The answers are the x-intercepts, so I just assumed that you (or the question) mixed things up. :smile:

    x-intercepts are the points where y=0; y-intercepts are where x=0. (The y-intercept of this function is (0, 10).)
     
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