how do you use y = kx(x-a)^2

[gritty frost]

If he can just sub -6 for a why can't he use -2?

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[edgy sapphire]

You cannot use -2 since it isn't a root, the roots are 0 and 6

We know that 6 is a root of double multiplicity because of the behavior of the curve C about 6, since it doesn't change sign at all

[gritty frost]

Ohhhhh the roots are where the graph cuts the x axis?

[edgy sapphire]

Yes, sometimes they are called x-intercepts

[gritty frost]

So every cubic graph has at least 1 root?

[edgy sapphire]

Yes.

[gritty frost]

Cool

[edgy sapphire]

Many have two or three even

No more than three, though

[gritty frost]

How did this guy just use kx(x-6)^2

[edgy sapphire]

So he noted that the polynomial is a cubic and it has an odd-multiplicity root at $x = 0$ and an even-multiplicity root at $x = 6$, so he obtains $y = k(x-0)^a(x-6)^b$ where $a$ is odd and $b$ is even, with $a,b>0$. We know that $a + b = 3$ and the only solutions which satisfy $a$ is odd and $b$ even is $a=1,b=2$. So we get $kx^1(x-6)^2 = kx(x-6)^2$.

[flat tangle]

John 10:34 ("Wannabe" John)

[gritty frost]

Is every cubic graph in the form of ax^3 + bx^2 + cx + d?

[edgy sapphire]

Yes

[gritty frost]

What do you mean odd multiplicy and even multiplicity?

[gritty frost]

So he put it in the form of k(x-x1)(x-x2)(x-x3)

But since there are only 2 roots you ignore x-x3?

[edgy sapphire]

Well, $x_3 = x_2$ essentially

[flat tangle]

John 10:34 ("Wannabe" John)

[edgy sapphire]

A root has a multiplicity of $k$ if it appears $k$ times in the factorization

[flat tangle]

John 10:34 ("Wannabe" John)

[edgy sapphire]

So $f(x) = (x-1)(x-1)(x+1)$, with this $x = 1$ has a multiplicity of $2$

[flat tangle]

John 10:34 ("Wannabe" John)

[edgy sapphire]

Whereas $x = -1$ has a multiplicity of $1$

[flat tangle]

John 10:34 ("Wannabe" John)

[gritty frost]

So if there's not a 3rd root than it's x2=x3

[edgy sapphire]

Yes

Well, only if there are 2 roots

[gritty frost]

Thank you very much