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\[y = 4x^3 -40x^2 + 125x -126\]

Can you take it from here?

yea

Great. Good luck. :)

thanks

You're welcome :)

i get \[y' = 12x^2 - 80x +125\]
what now

im supposed to get 3 slopes

factor right?

\[y = (2x-5)(6x-25)\]

y prime i mean

slopes so far are 5/2, 25/6, and...?

Just a sec iop360...

actually is that even right?

ok

Now, what we want to do is first get the positions where the tangents touch the curve...

Think about it...how would we do that?

Hint: Simultaneous equations.

equate the two and solve for x?

Sorry for taking forever to reply...

\[\frac{ (4x^3 - 40x^2 +125x -126) - (-1) }{ x-1 }= 12x^2 -80x + 125\]

its alright

we would have to find m first

True. Trying something different...

i think i found the slopes, tell me what you have too
slope = 0, 5/2, 4

ok

you mean now find the line equations?

Yep.

y = -1
y = (5/2)x - (7/2)
y = 4x - 5

doesnt seem right when i graph it...

hopefully u have a better one

Well, I've graphed yours and It seems okay. :) My equation is giving me imaginary solutions -_-

u graphed mine?

Yep.

it doesnt look tangent to the curve

Plotting on a calculator?

yes

they just all intersect together and dont touch the curve

sure

perhaps i should bump this

i see

alright

ill leave this question open

thanks for helping

plug in the found slopes in point-slope formula and it should be right

Nice. Good job! :)