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

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- anonymous

\[y = 4x^3 -40x^2 + 125x -126\]

- 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

Can you take it from here?

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## More answers

- anonymous

yea

- allank

Great. Good luck. :)

- anonymous

thanks

- allank

You're welcome :)

- anonymous

i get \[y' = 12x^2 - 80x +125\]
what now

- anonymous

im supposed to get 3 slopes

- anonymous

factor right?

- anonymous

\[y = (2x-5)(6x-25)\]

- anonymous

y prime i mean

- anonymous

slopes so far are 5/2, 25/6, and...?

- allank

Just a sec iop360...

- anonymous

actually is that even right?

- anonymous

ok

- allank

Now, what we want to do is first get the positions where the tangents touch the curve...

- allank

Think about it...how would we do that?

- allank

Hint: Simultaneous equations.

- anonymous

so use m = m
with one m representing the first derivative i just calculated
and the other m representing rise/run

- anonymous

equate the two and solve for x?

- allank

Sorry for taking forever to reply...

- anonymous

\[\frac{ (4x^3 - 40x^2 +125x -126) - (-1) }{ x-1 }= 12x^2 -80x + 125\]

- anonymous

its alright

- 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

we would have to find m first

- allank

True. Trying something different...

- anonymous

i think i found the slopes, tell me what you have too
slope = 0, 5/2, 4

- 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

ok

- anonymous

you mean now find the line equations?

- allank

Yep.

- anonymous

k

- anonymous

y = -1
y = (5/2)x - (7/2)
y = 4x - 5

- anonymous

doesnt seem right when i graph it...

- anonymous

hopefully u have a better one

- allank

Well, I've graphed yours and It seems okay. :) My equation is giving me imaginary solutions -_-

- anonymous

u graphed mine?

- allank

Yep.

- anonymous

it doesnt look tangent to the curve

- allank

Plotting on a calculator?

- anonymous

yes

- anonymous

they just all intersect together and dont touch the curve

- 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

sure

- anonymous

perhaps i should bump this

- 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

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

i see

- 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

alright

- anonymous

ill leave this question open

- anonymous

thanks for helping

- 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

plug in the found slopes in point-slope formula and it should be right

- allank

Nice. Good job! :)

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