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kirk.freedman

  • 3 years ago

The rational function has a y-intercept of 7. What is the equation for this function?

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  1. kirk.freedman
    • 3 years ago
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  2. kirk.freedman
    • 3 years ago
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    @phi @UnkleRhaukus @robtobey @Mertsj @mathstudent55 @jim_thompson5910 @ghazi

  3. ghazi
    • 3 years ago
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    just shift the origin of hyperbola by (-2,5)

  4. kirk.freedman
    • 3 years ago
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    I'm sorry, I don't really understand.

  5. kirk.freedman
    • 3 years ago
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    Somebody please please help!

  6. tkhunny
    • 3 years ago
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    1) Vertical Asympote @ x = -2 ==> \(\dfrac{Something}{x+2}\) 2) Horizontal Asympote @ y = 5 ==> \(\dfrac{5x - Something}{x+2}\) 3) y-intercept (0,7) ==> -Something / 2 = 7 ==> \(\dfrac{5x + 14}{x+2}\) 4) Almost. We missed (-3,0). How do you suppose we should alter it for that?

  7. kirk.freedman
    • 3 years ago
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    I'm sorry, I would give it a guess but I really don't know where we would add that

  8. kirk.freedman
    • 3 years ago
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    what is the (-3,0)? is there a name for it like the others?

  9. tkhunny
    • 3 years ago
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    x-intercept

  10. kirk.freedman
    • 3 years ago
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    Oh Oh Oh, I am actually understanding this, and I was gonna guess x-int, but I thought it was wrong.. So I understand everything, but I just don't know how you can put the x-int into the equation :)

  11. tkhunny
    • 3 years ago
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    A little more generally... General Form \(\dfrac{ax-b}{cx-d}\) 1) Vertical Asymptote @ x = -2 ==> c(-2) - d = 0 ==> d = -2c General Form \(\dfrac{ax-b}{cx-(-2c)} = \dfrac{ax-b}{cx + 2c} = \dfrac{ax-b}{c(x + 2)}\) 2) Horizontal Asymptote @ y = 5 ==> a = 5c General Form \(\dfrac{5cx-b}{c(x + 2)}\) 3) y-intercept (0,7) ==> \(\dfrac{-b}{2c} = 7\) ==> \(b = -14c\) General Form \(\dfrac{5cx+14c}{c(x + 2)} = \dfrac{c(5x+14)}{c(x + 2)} = \dfrac{5x+14}{x + 2}\). This is very unfortunate. It means we must have the original form wrong onless the last piece just happens to work as desired. 4) x-intercept (-3,0) ==> \(\dfrac{5(-3)+14}{(-3) + 2} = \dfrac{-15+14}{-3+2} = \dfrac{-1}{-1} = 1\) And we were hoping for zero.

  12. kirk.freedman
    • 3 years ago
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    Ok, so if it doesn't work out how do I answer it?

  13. kirk.freedman
    • 3 years ago
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    So would the equation for the function be (5x+14)/(x+2)?

  14. tkhunny
    • 3 years ago
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    Nope. Doesn't quite work. It needs some reshaping. \(y = \dfrac{5(x+3)}{x+2}\) hits (-3,0) but not (0,7). Very annoying. I have a plan. Let's play like the x-intercept is only CLOSE to (-3,0). The problem statement didn't actually say that, so maybe it's just an approximate value. In fact, (-3,1) looks better than (-3,0), so maybe we ARE done! \(y = \dfrac{5x+14}{x+2}\) it is!!!

  15. kirk.freedman
    • 3 years ago
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    Thanks, I appreciate it sooo much! I know how tough it is to type all of this out, but I am sure this will help others too in the future!

  16. tkhunny
    • 3 years ago
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    We probably should note the x-intercept on which we settled. 5x + 14 = 0 ==> x = -14/5 ==> (-14/5, 0). Done. There is actually a little more information in the nature of the vertical asymptote. Maybe we'll run into one of those, later.

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