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

- anonymous

- schrodinger

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- anonymous

##### 1 Attachment

- anonymous

\[h(x)=k(x+3)(x-2)(x-3)\]

- anonymous

3=k(2+3)(2-2)(2-3)

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## More answers

- anonymous

3=k

- anonymous

Looks alright so far.

- anonymous

is there more to or is it all ? :S

- anonymous

Er., wait, no

- anonymous

In your solving for k, you need to put in 0 for x if you're using the y-intercept.

- anonymous

would that make k zero ?

- anonymous

No, what you did made 3=0.

- anonymous

ohh i see

- anonymous

"3=k(2+3)(2-2)(2-3)"
3=k(5)(0)(-1) -> 3=0

- anonymous

i cancelled out k too :/

- anonymous

(2,3) is not a solution to the equation, but (0,3) is.

- anonymous

is dont see where you're getting (2,3) from

- anonymous

You plugged in x=2 and set y=3 in your attempt to solve for k.

- anonymous

So, I don't see where *you* got (2,3) from. ;-)

- anonymous

ohhhh ! i'm supposed to set x as 0 right ?

- anonymous

Yes, because you know the y-intercept is 3.

- anonymous

ohh okay i see it now :$

- anonymous

You could use any other known point, but that one is the most convenient.

- anonymous

so k would be 1/6

- anonymous

and h(x) = 1/6 (x+3)(x-2)(x-3)

- anonymous

never mind
h(x) = k(x+3)(x-2)(x-3)
h(0)=3
3=k(3)(-2)(-3)
3=18k

- anonymous

Yep, it's all good.

- anonymous

Thank-you once again :D

- anonymous

if you're not busy, can you help me with one last question

- anonymous

I think you know what you're doing, just be careful and don't rush through it.

- anonymous

Oh, sure. I think I can handle one more.

- anonymous

thanks :)

- anonymous

Find the equation of a cubic polynomial that gives a remainder of 3, when divided by (x+2)

- anonymous

how would i approach this problem ?

- anonymous

Do you remember the remainder theorem?

- anonymous

yes, the factor is equal to zero

- anonymous

Sort of..

- 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

but how would i use this info to make an equation, this is like working backwards from the norm :/

- anonymous

Yes, a lot of what you're doing here is using concepts to 'unsolve' equations.

- anonymous

If dividing by (x+2) gives a remainder of 3, then you know (-2,3) is a solution to the equation.

- anonymous

yes that's true

- 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

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

It wants a cubic function, so what does an equation for one of those look like?

- anonymous

like x^3-3x^2-4x-5

- anonymous

Sure, something like that, so to be more general, start with
\[y=ax^3+bx^2+cx+d\]

- 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

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

so something like \[y=x^3+x^2+x+d \] ?

- anonymous

I don't see why not.

- anonymous

hmm but "d" has to be a specific number right

- anonymous

Yep, that's what you're solving for.
Me, I went with y=ax^3, then solved for a.

- anonymous

(but I'm lazy like that)

- anonymous

ohh that way a would be -3/8 right

- anonymous

Uh huh.

- 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

\[:. y=\frac{ -3 }{ 8 }x^3\]

- anonymous

Looks beautiful.

- anonymous

thank you so much for your helppp, you're amazing :D

- anonymous

My pleasure.

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