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anonymous
 3 years ago
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)
anonymous
 3 years ago
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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y = 4x^3 40x^2 + 125x 126\]

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Cool question. I'd start by calculating the first derivative of y which will give me the slope of the lines we are looking for.

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Can you take it from here?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i get \[y' = 12x^2  80x +125\] what now

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im supposed to get 3 slopes

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y = (2x5)(6x25)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0slopes so far are 5/2, 25/6, and...?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually is that even right?

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Now, what we want to do is first get the positions where the tangents touch the curve...

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Think about it...how would we do that?

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Hint: Simultaneous equations.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so use m = m with one m representing the first derivative i just calculated and the other m representing rise/run

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0equate the two and solve for x?

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry for taking forever to reply...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ (4x^3  40x^2 +125x 126)  (1) }{ x1 }= 12x^2 80x + 125\]

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, we have the gradient: m = 12x^280x+125 and we have a point (1,1). I'm thinking we can use the general form of an equation of a line: yy1 = m(xx1) where y1 and x1 are 1 and 1 respectively, and m is the gradient. I'm not sure whether that will work. Lemme try....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we would have to find m first

allank
 3 years ago
Best ResponseYou've already chosen the best response.0True. Trying something different...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think i found the slopes, tell me what you have too slope = 0, 5/2, 4

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you mean now find the line equations?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y = 1 y = (5/2)x  (7/2) y = 4x  5

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0doesnt seem right when i graph it...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hopefully u have a better one

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Well, I've graphed yours and It seems okay. :) My equation is giving me imaginary solutions _

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it doesnt look tangent to the curve

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Plotting on a calculator?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0they just all intersect together and dont touch the curve

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Darn. Okay, maybe I'm just sleepy to notice that on a computer grapher...so wanna see my method? Though it may not lead to an answer...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0perhaps i should bump this

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, so we have the slope m = 12x^280x+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(x1) Plugging in m: y = m(x1)1 y = (12x^280x+125)(x1) 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)

allank
 3 years ago
Best ResponseYou've already chosen the best response.0I 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*

allank
 3 years ago
Best ResponseYou've already chosen the best response.0Do 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ill leave this question open

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hey, 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0plug in the found slopes in pointslope formula and it should be right
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