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 2 years ago
The rational function has a yintercept of 7. What is the equation for this function?
 2 years ago
The rational function has a yintercept of 7. What is the equation for this function?

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kirk.freedman
 2 years ago
Best ResponseYou've already chosen the best response.0@phi @UnkleRhaukus @robtobey @Mertsj @mathstudent55 @jim_thompson5910 @ghazi

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.1just shift the origin of hyperbola by (2,5)

kirk.freedman
 2 years ago
Best ResponseYou've already chosen the best response.0I'm sorry, I don't really understand.

kirk.freedman
 2 years ago
Best ResponseYou've already chosen the best response.0Somebody please please help!

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.11) Vertical Asympote @ x = 2 ==> \(\dfrac{Something}{x+2}\) 2) Horizontal Asympote @ y = 5 ==> \(\dfrac{5x  Something}{x+2}\) 3) yintercept (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?

kirk.freedman
 2 years ago
Best ResponseYou've already chosen the best response.0I'm sorry, I would give it a guess but I really don't know where we would add that

kirk.freedman
 2 years ago
Best ResponseYou've already chosen the best response.0what is the (3,0)? is there a name for it like the others?

kirk.freedman
 2 years ago
Best ResponseYou've already chosen the best response.0Oh Oh Oh, I am actually understanding this, and I was gonna guess xint, but I thought it was wrong.. So I understand everything, but I just don't know how you can put the xint into the equation :)

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1A little more generally... General Form \(\dfrac{axb}{cxd}\) 1) Vertical Asymptote @ x = 2 ==> c(2)  d = 0 ==> d = 2c General Form \(\dfrac{axb}{cx(2c)} = \dfrac{axb}{cx + 2c} = \dfrac{axb}{c(x + 2)}\) 2) Horizontal Asymptote @ y = 5 ==> a = 5c General Form \(\dfrac{5cxb}{c(x + 2)}\) 3) yintercept (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) xintercept (3,0) ==> \(\dfrac{5(3)+14}{(3) + 2} = \dfrac{15+14}{3+2} = \dfrac{1}{1} = 1\) And we were hoping for zero.

kirk.freedman
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, so if it doesn't work out how do I answer it?

kirk.freedman
 2 years ago
Best ResponseYou've already chosen the best response.0So would the equation for the function be (5x+14)/(x+2)?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Nope. 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 xintercept 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!!!

kirk.freedman
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks, 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!

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1We probably should note the xintercept 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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