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iop360
Group Title
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)
 2 years ago
 2 years ago
iop360 Group Title
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)
 2 years ago
 2 years ago

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iop360 Group TitleBest ResponseYou've already chosen the best response.2
\[y = 4x^3 40x^2 + 125x 126\]
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

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

allank Group TitleBest ResponseYou've already chosen the best response.0
Great. Good luck. :)
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
You're welcome :)
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
i get \[y' = 12x^2  80x +125\] what now
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
im supposed to get 3 slopes
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
factor right?
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
\[y = (2x5)(6x25)\]
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
y prime i mean
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
slopes so far are 5/2, 25/6, and...?
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
Just a sec iop360...
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
actually is that even right?
 2 years ago

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

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

allank Group TitleBest ResponseYou've already chosen the best response.0
Hint: Simultaneous equations.
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
so use m = m with one m representing the first derivative i just calculated and the other m representing rise/run
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
equate the two and solve for x?
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
Sorry for taking forever to reply...
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
\[\frac{ (4x^3  40x^2 +125x 126)  (1) }{ x1 }= 12x^2 80x + 125\]
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
Okay, 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....
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
we would have to find m first
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
True. Trying something different...
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
i think i found the slopes, tell me what you have too slope = 0, 5/2, 4
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
you mean now find the line equations?
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
y = 1 y = (5/2)x  (7/2) y = 4x  5
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
doesnt seem right when i graph it...
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
hopefully u have a better one
 2 years ago

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

iop360 Group TitleBest ResponseYou've already chosen the best response.2
u graphed mine?
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
it doesnt look tangent to the curve
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
Plotting on a calculator?
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
they just all intersect together and dont touch the curve
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
Darn. 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...
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
perhaps i should bump this
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
Okay, 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)
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
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*
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
ill leave this question open
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
thanks for helping
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
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
 2 years ago

iop360 Group TitleBest ResponseYou've already chosen the best response.2
plug in the found slopes in pointslope formula and it should be right
 2 years ago

allank Group TitleBest ResponseYou've already chosen the best response.0
Nice. Good job! :)
 2 years ago
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