anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[y = 4x^3 -40x^2 + 125x -126\]
allank
  • allank
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.
allank
  • allank
Can you take it from here?

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anonymous
  • anonymous
yea
allank
  • allank
Great. Good luck. :)
anonymous
  • anonymous
thanks
allank
  • allank
You're welcome :)
anonymous
  • anonymous
i get \[y' = 12x^2 - 80x +125\] what now
anonymous
  • anonymous
im supposed to get 3 slopes
anonymous
  • anonymous
factor right?
anonymous
  • anonymous
\[y = (2x-5)(6x-25)\]
anonymous
  • anonymous
y prime i mean
anonymous
  • anonymous
slopes so far are 5/2, 25/6, and...?
allank
  • allank
Just a sec iop360...
anonymous
  • anonymous
actually is that even right?
anonymous
  • anonymous
ok
allank
  • allank
Now, what we want to do is first get the positions where the tangents touch the curve...
allank
  • allank
Think about it...how would we do that?
allank
  • allank
Hint: Simultaneous equations.
anonymous
  • anonymous
so use m = m with one m representing the first derivative i just calculated and the other m representing rise/run
anonymous
  • anonymous
equate the two and solve for x?
allank
  • allank
Sorry for taking forever to reply...
anonymous
  • anonymous
\[\frac{ (4x^3 - 40x^2 +125x -126) - (-1) }{ x-1 }= 12x^2 -80x + 125\]
anonymous
  • anonymous
its alright
allank
  • allank
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....
anonymous
  • anonymous
we would have to find m first
allank
  • allank
True. Trying something different...
anonymous
  • anonymous
i think i found the slopes, tell me what you have too slope = 0, 5/2, 4
allank
  • allank
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?
anonymous
  • anonymous
ok
anonymous
  • anonymous
you mean now find the line equations?
allank
  • allank
Yep.
anonymous
  • anonymous
k
anonymous
  • anonymous
y = -1 y = (5/2)x - (7/2) y = 4x - 5
anonymous
  • anonymous
doesnt seem right when i graph it...
anonymous
  • anonymous
hopefully u have a better one
allank
  • allank
Well, I've graphed yours and It seems okay. :) My equation is giving me imaginary solutions -_-
anonymous
  • anonymous
u graphed mine?
allank
  • allank
Yep.
anonymous
  • anonymous
it doesnt look tangent to the curve
allank
  • allank
Plotting on a calculator?
anonymous
  • anonymous
yes
anonymous
  • anonymous
they just all intersect together and dont touch the curve
allank
  • allank
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...
anonymous
  • anonymous
sure
anonymous
  • anonymous
perhaps i should bump this
allank
  • allank
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)
allank
  • allank
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*
anonymous
  • anonymous
i see
allank
  • allank
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.
anonymous
  • anonymous
alright
anonymous
  • anonymous
ill leave this question open
anonymous
  • anonymous
thanks for helping
anonymous
  • anonymous
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
anonymous
  • anonymous
plug in the found slopes in point-slope formula and it should be right
allank
  • allank
Nice. Good job! :)

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