burhan101
find the equation of this graph (picture included) i have started it off but don't know how to finish it
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burhan101
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burhan101
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\[h(x)=k(x+3)(x-2)(x-3)\]
burhan101
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3=k(2+3)(2-2)(2-3)
burhan101
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3=k
CliffSedge
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Looks alright so far.
burhan101
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is there more to or is it all ? :S
CliffSedge
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Er., wait, no
CliffSedge
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In your solving for k, you need to put in 0 for x if you're using the y-intercept.
burhan101
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would that make k zero ?
CliffSedge
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No, what you did made 3=0.
burhan101
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ohh i see
CliffSedge
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"3=k(2+3)(2-2)(2-3)"
3=k(5)(0)(-1) -> 3=0
burhan101
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i cancelled out k too :/
CliffSedge
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(2,3) is not a solution to the equation, but (0,3) is.
burhan101
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is dont see where you're getting (2,3) from
CliffSedge
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You plugged in x=2 and set y=3 in your attempt to solve for k.
CliffSedge
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So, I don't see where *you* got (2,3) from. ;-)
burhan101
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ohhhh ! i'm supposed to set x as 0 right ?
CliffSedge
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Yes, because you know the y-intercept is 3.
burhan101
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ohh okay i see it now :$
CliffSedge
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You could use any other known point, but that one is the most convenient.
burhan101
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so k would be 1/6
burhan101
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and h(x) = 1/6 (x+3)(x-2)(x-3)
CliffSedge
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never mind
h(x) = k(x+3)(x-2)(x-3)
h(0)=3
3=k(3)(-2)(-3)
3=18k
CliffSedge
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Yep, it's all good.
burhan101
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Thank-you once again :D
burhan101
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if you're not busy, can you help me with one last question
CliffSedge
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I think you know what you're doing, just be careful and don't rush through it.
CliffSedge
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Oh, sure. I think I can handle one more.
burhan101
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thanks :)
burhan101
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Find the equation of a cubic polynomial that gives a remainder of 3, when divided by (x+2)
burhan101
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how would i approach this problem ?
CliffSedge
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Do you remember the remainder theorem?
burhan101
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yes, the factor is equal to zero
CliffSedge
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Sort of..
CliffSedge
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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.
burhan101
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but how would i use this info to make an equation, this is like working backwards from the norm :/
CliffSedge
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Yes, a lot of what you're doing here is using concepts to 'unsolve' equations.
CliffSedge
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If dividing by (x+2) gives a remainder of 3, then you know (-2,3) is a solution to the equation.
burhan101
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yes that's true
CliffSedge
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It's a pretty open-ended question; there will be many functions (infinite, really) that you could develop that have that point on it.
burhan101
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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 :/
CliffSedge
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It wants a cubic function, so what does an equation for one of those look like?
burhan101
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like x^3-3x^2-4x-5
CliffSedge
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Sure, something like that, so to be more general, start with
\[y=ax^3+bx^2+cx+d\]
CliffSedge
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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.
CliffSedge
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You could even set three of them =0 if you wanted to be really lazy about it.
(General math tip: be lazy whenever possible)
burhan101
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so something like \[y=x^3+x^2+x+d \] ?
CliffSedge
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I don't see why not.
burhan101
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hmm but "d" has to be a specific number right
CliffSedge
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Yep, that's what you're solving for.
Me, I went with y=ax^3, then solved for a.
CliffSedge
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(but I'm lazy like that)
burhan101
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ohh that way a would be -3/8 right
CliffSedge
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Uh huh.
CliffSedge
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And if you want, you can verify that
\[\large\frac{-3}{8}x^3 \div (x+2)\]
will give a remainder of 3.
burhan101
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\[:. y=\frac{ -3 }{ 8 }x^3\]
CliffSedge
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Looks beautiful.
burhan101
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thank you so much for your helppp, you're amazing :D
CliffSedge
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My pleasure.