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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)
 one year ago
 one year 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)
 one year ago
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

iop360 Group TitleBest ResponseYou've already chosen the best response.2
its alright
 one year 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....
 one year ago

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

allank Group TitleBest ResponseYou've already chosen the best response.0
True. Trying something different...
 one year 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
 one year 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?
 one year ago

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

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

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

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

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

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

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

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

iop360 Group TitleBest ResponseYou've already chosen the best response.2
perhaps i should bump this
 one year 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)
 one year 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*
 one year 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.
 one year ago

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

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

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