Calculus Midoriさんのイラストまとめ

>We're nearly there! (Woohoo!)
>Now we just need to do the finishing touches before we off and frame it in a gallery or somewhere.
>With our current description, both images here are valid.

>Gosh, I'm sure Keiji's buddy wished he was raised to an even power. Seeing as he didn't bounce back up when his body was pelted like swiss cheese, I'm awful sure he's some linear factor to an odd power.
>He simply crossed from the realm of the living... to the realm of the dead.

>Going back to our example, that means at x=-3, the line crosses. There's an invisible power of "1" (odd) that we don't write for redundancy, but it's there and beating!
>The same goes for x=-1, it crosses, while x=2 bounces!

>Zeros A and B have a linear factor to an odd power.
>Zero C has a linear factor to an even number.
>Got it? Got it. I'm moving along now.
>I want to finish this already too, you know. I need to get back to watching the participants before my boredom kills me a second time.

>These "bounces" and "crosses" are determined by the multiplicity, again, I say. Is it me or is being a doll for so long making me sound like a broken record? Haha.
>Linear factors to an odd power will "cross".
>Linear factors to an even power will "bounce".

>So far, we know that the leading coefficient's power is even and higher than three, and that there are 3 zeros.
>We're not done yet! We can also determine the powers of the zeros as well.
>At A and B, the line crosses the x-axis, while at C, it bounces.

>Why, why... you ask?
>Both of the sides are approaching towards positive infinity (y-value) as the x-value approaches negative AND positive infinity. They have the same end behavior.
>If your memory's swell, which it should be, that means the leading coefficient's power is even.

>It intersects with the x-axis 3 times, at the points I've labeled point A, B, and C.
>We can assume that the power of the leading coefficient is 3 or at least 3.
>Though, we know that it can't possibly be 3, just like how this death game can't possibly be a bad dream!

>I don't want to post the graph of our lovely example too soon, It'll spoil the fun. I have a short figure for this explanation.
>We don't have the exact equation, but based on the graph's shape, we can assume so much.

>The "it" is the zeros. We determine the multiplicity by looking at the power of the linear factors.
>For (x-2)^2, (x-2) is the linear factor.
>We won't take a look at the magnitude, we'll only take a look at whether it's even or odd, black or white, dead or alive.