Question

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

Answer 1

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

Answer 2

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

Answer 3

3=k

Answer 4

Looks alright so far.

Answer 5

is there more to or is it all ? :S

Answer 6

Er., wait, no

Answer 7

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

Answer 8

would that make k zero ?

Answer 9

No, what you did made 3=0.

Answer 10

ohh i see

Answer 11

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

3=k(5)(0)(-1) -> 3=0

Answer 12

i cancelled out k too :/

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

ohh okay i see it now :$

Answer 20

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

Answer 21

so k would be 1/6

Answer 22

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

Answer 23

never mind

h(x) = k(x+3)(x-2)(x-3)
h(0)=3
3=k(3)(-2)(-3)
3=18k

Answer 24

Yep, it's all good.

Answer 25

Thank-you once again :D

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

thanks :)

Answer 30

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

Answer 31

how would i approach this problem ?

Answer 32

Do you remember the remainder theorem?

Answer 33

yes, the factor is equal to zero

Answer 34

Sort of..

Answer 35

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.

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

yes that's true

Answer 40

It's a pretty open-ended question; there will be many functions (infinite, really) that you could develop that have that point on it.

Answer 41

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 :/

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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.

Answer 46

You could even set three of them =0 if you wanted to be really lazy about it.
(General math tip: be lazy whenever possible)

Answer 47

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

Answer 48

I don't see why not.

Answer 49

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

Answer 50

Yep, that's what you're solving for.

Me, I went with y=ax^3, then solved for a.

Answer 51

(but I'm lazy like that)

Answer 52

ohh that way a would be -3/8 right

Answer 53

Uh huh.

Answer 54

And if you want, you can verify that
\[\large\frac{-3}{8}x^3 \div (x+2)\]
will give a remainder of 3.

Answer 55

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

Answer 56

Looks beautiful.

Answer 57

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

Answer 58

My pleasure.