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find the equations of three tangent lines to the graph of y = 4x^3 -40x^2 + 125x - 126 which pass through point (1, -1) (but not on the curve)

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\[y = 4x^3 -40x^2 + 125x -126\]
Cool question. I'd start by calculating the first derivative of y which will give me the slope of the lines we are looking for.
Can you take it from here?

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Great. Good luck. :)
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?
Now, what we want to do is first get the positions where the tangents touch the curve...
Think about would we do that?
Hint: Simultaneous equations.
so use m = m with one m representing the first derivative i just calculated and the other m representing rise/run
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
Okay, we have the gradient: m = 12x^2-80x+125 and we have a point (1,-1). I'm thinking we can use the general form of an equation of a line: y-y1 = m(x-x1) where y1 and x1 are -1 and 1 respectively, and m is the gradient. I'm not sure whether that will work. Lemme try....
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
Well, I have used a different method, where I first determine the positions where the tangents touch the curve. Want to finish off and compare answers?
you mean now find the line equations?
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?
it doesnt look tangent to the curve
Plotting on a calculator?
they just all intersect together and dont touch the curve
Darn. Okay, maybe I'm just sleepy to notice that on a computer wanna see my method? Though it may not lead to an answer...
perhaps i should bump this
Okay, so we have the slope m = 12x^2-80x+125. Remember that the value of m depends on the x value on the curve.... Knowing the point (1,-1) , the equation of the tangent(s) will be y+1 = m(x-1) Plugging in m: y = m(x-1)-1 y = (12x^2-80x+125)(x-1) -1 y = 12x^3 - 80x^2 +125x - 12x^2 + 80x -125 - 1 y = 12x^3 - 92x^2 + 205x - 126 (equation 1) So, we have an equation that depends on the very same x value on the curve as the original equation: y = 4x^3 -40x^2 + 125x - 126 (equation 2) Two nice simultaneous equations that will give the x values on the curve where the tangents touch the curve: y = 12x^3 - 92x^2 + 205x - 126 (equation 1) y = 4x^3 -40x^2 + 125x - 126 (equation 2)
I hope I've done the math right. Subtracting the second from the first eliminates y and the 126. y = 12x^3 - 92x^2 + 205x - 126 (equation 1) y = 4x^3 -40x^2 + 125x - 126 (equation 2) 0 = 8x^3 +52x^2 + 330x And solving for x gives x=0 and two other imaginary solutions. *sigh*
i see
Do bump @iop360. I'll sleep on it and in case you haven't gotten a valid response by the time I'm awake, I will post the solution.
ill leave this question open
thanks for helping
hey, i got the question right! if you want to know how, the values for x i found (0, (5/2), and 4) were actually the x coordinates of the tangents so i plugged these values in the original equation to get their respective y values after that i used each (x,y) coordinates of the three tangents with the given point in the question (1,-1) in the rise/run formula to get the slope for each line
plug in the found slopes in point-slope formula and it should be right
Nice. Good job! :)

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