anonymous
  • anonymous
find the equation of this graph (picture included) i have started it off but don't know how to finish it
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
\[h(x)=k(x+3)(x-2)(x-3)\]
anonymous
  • anonymous
3=k(2+3)(2-2)(2-3)

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anonymous
  • anonymous
3=k
anonymous
  • anonymous
Looks alright so far.
anonymous
  • anonymous
is there more to or is it all ? :S
anonymous
  • anonymous
Er., wait, no
anonymous
  • anonymous
In your solving for k, you need to put in 0 for x if you're using the y-intercept.
anonymous
  • anonymous
would that make k zero ?
anonymous
  • anonymous
No, what you did made 3=0.
anonymous
  • anonymous
ohh i see
anonymous
  • anonymous
"3=k(2+3)(2-2)(2-3)" 3=k(5)(0)(-1) -> 3=0
anonymous
  • anonymous
i cancelled out k too :/
anonymous
  • anonymous
(2,3) is not a solution to the equation, but (0,3) is.
anonymous
  • anonymous
is dont see where you're getting (2,3) from
anonymous
  • anonymous
You plugged in x=2 and set y=3 in your attempt to solve for k.
anonymous
  • anonymous
So, I don't see where *you* got (2,3) from. ;-)
anonymous
  • anonymous
ohhhh ! i'm supposed to set x as 0 right ?
anonymous
  • anonymous
Yes, because you know the y-intercept is 3.
anonymous
  • anonymous
ohh okay i see it now :$
anonymous
  • anonymous
You could use any other known point, but that one is the most convenient.
anonymous
  • anonymous
so k would be 1/6
anonymous
  • anonymous
and h(x) = 1/6 (x+3)(x-2)(x-3)
anonymous
  • anonymous
never mind h(x) = k(x+3)(x-2)(x-3) h(0)=3 3=k(3)(-2)(-3) 3=18k
anonymous
  • anonymous
Yep, it's all good.
anonymous
  • anonymous
Thank-you once again :D
anonymous
  • anonymous
if you're not busy, can you help me with one last question
anonymous
  • anonymous
I think you know what you're doing, just be careful and don't rush through it.
anonymous
  • anonymous
Oh, sure. I think I can handle one more.
anonymous
  • anonymous
thanks :)
anonymous
  • anonymous
Find the equation of a cubic polynomial that gives a remainder of 3, when divided by (x+2)
anonymous
  • anonymous
how would i approach this problem ?
anonymous
  • anonymous
Do you remember the remainder theorem?
anonymous
  • anonymous
yes, the factor is equal to zero
anonymous
  • anonymous
Sort of..
anonymous
  • anonymous
It kind of goes like this. If (x+2) is a factor of the polynomial, then dividing by (x+2) would have a remainder of 0. And, if (x+2) is a factor, then x = -2 is a root of the polynomial. If (x+2) is not a factor, then f(-2) = the remainder.
anonymous
  • anonymous
but how would i use this info to make an equation, this is like working backwards from the norm :/
anonymous
  • anonymous
Yes, a lot of what you're doing here is using concepts to 'unsolve' equations.
anonymous
  • anonymous
If dividing by (x+2) gives a remainder of 3, then you know (-2,3) is a solution to the equation.
anonymous
  • anonymous
yes that's true
anonymous
  • anonymous
It's a pretty open-ended question; there will be many functions (infinite, really) that you could develop that have that point on it.
anonymous
  • anonymous
i still don't really get it, like i understand what it is asking for but i don't knw how to get to it :/
anonymous
  • anonymous
It wants a cubic function, so what does an equation for one of those look like?
anonymous
  • anonymous
like x^3-3x^2-4x-5
anonymous
  • anonymous
Sure, something like that, so to be more general, start with \[y=ax^3+bx^2+cx+d\]
anonymous
  • anonymous
You know what x and y are, and three out of those four variables are free parameters, so choose whatever you want for those, then solve for the fourth one.
anonymous
  • anonymous
You could even set three of them =0 if you wanted to be really lazy about it. (General math tip: be lazy whenever possible)
anonymous
  • anonymous
so something like \[y=x^3+x^2+x+d \] ?
anonymous
  • anonymous
I don't see why not.
anonymous
  • anonymous
hmm but "d" has to be a specific number right
anonymous
  • anonymous
Yep, that's what you're solving for. Me, I went with y=ax^3, then solved for a.
anonymous
  • anonymous
(but I'm lazy like that)
anonymous
  • anonymous
ohh that way a would be -3/8 right
anonymous
  • anonymous
Uh huh.
anonymous
  • anonymous
And if you want, you can verify that \[\large\frac{-3}{8}x^3 \div (x+2)\] will give a remainder of 3.
anonymous
  • anonymous
\[:. y=\frac{ -3 }{ 8 }x^3\]
anonymous
  • anonymous
Looks beautiful.
anonymous
  • anonymous
thank you so much for your helppp, you're amazing :D
anonymous
  • anonymous
My pleasure.

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