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find the equation of this graph (picture included) i have started it off but don't know how to finish it
 one year ago
 one year ago
find the equation of this graph (picture included) i have started it off but don't know how to finish it
 one year ago
 one year ago

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burhan101Best ResponseYou've already chosen the best response.1
\[h(x)=k(x+3)(x2)(x3)\]
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Looks alright so far.
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
is there more to or is it all ? :S
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
In your solving for k, you need to put in 0 for x if you're using the yintercept.
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
would that make k zero ?
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
No, what you did made 3=0.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
"3=k(2+3)(22)(23)" 3=k(5)(0)(1) > 3=0
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
i cancelled out k too :/
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
(2,3) is not a solution to the equation, but (0,3) is.
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
is dont see where you're getting (2,3) from
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
You plugged in x=2 and set y=3 in your attempt to solve for k.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
So, I don't see where *you* got (2,3) from. ;)
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
ohhhh ! i'm supposed to set x as 0 right ?
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Yes, because you know the yintercept is 3.
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
ohh okay i see it now :$
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
You could use any other known point, but that one is the most convenient.
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
and h(x) = 1/6 (x+3)(x2)(x3)
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
never mind h(x) = k(x+3)(x2)(x3) h(0)=3 3=k(3)(2)(3) 3=18k
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Yep, it's all good.
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
Thankyou once again :D
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
if you're not busy, can you help me with one last question
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
I think you know what you're doing, just be careful and don't rush through it.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Oh, sure. I think I can handle one more.
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
Find the equation of a cubic polynomial that gives a remainder of 3, when divided by (x+2)
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
how would i approach this problem ?
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Do you remember the remainder theorem?
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
yes, the factor is equal to zero
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
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.
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
but how would i use this info to make an equation, this is like working backwards from the norm :/
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Yes, a lot of what you're doing here is using concepts to 'unsolve' equations.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
If dividing by (x+2) gives a remainder of 3, then you know (2,3) is a solution to the equation.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
It's a pretty openended question; there will be many functions (infinite, really) that you could develop that have that point on it.
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
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 :/
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
It wants a cubic function, so what does an equation for one of those look like?
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Sure, something like that, so to be more general, start with \[y=ax^3+bx^2+cx+d\]
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
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.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
You could even set three of them =0 if you wanted to be really lazy about it. (General math tip: be lazy whenever possible)
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
so something like \[y=x^3+x^2+x+d \] ?
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
I don't see why not.
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
hmm but "d" has to be a specific number right
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Yep, that's what you're solving for. Me, I went with y=ax^3, then solved for a.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
(but I'm lazy like that)
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
ohh that way a would be 3/8 right
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
And if you want, you can verify that \[\large\frac{3}{8}x^3 \div (x+2)\] will give a remainder of 3.
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
\[:. y=\frac{ 3 }{ 8 }x^3\]
 one year ago

burhan101Best ResponseYou've already chosen the best response.1
thank you so much for your helppp, you're amazing :D
 one year ago
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