Question

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)

Answer 1

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

Answer 2

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.

Answer 3

Can you take it from here?

Answer 4

yea

Answer 5

Great. Good luck. :)

Answer 6

thanks

Answer 7

You're welcome :)

Answer 8

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

Answer 9

im supposed to get 3 slopes

Answer 10

factor right?

Answer 11

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

Answer 12

y prime i mean

Answer 13

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

Answer 14

Just a sec iop360...

Answer 15

actually is that even right?

Answer 16

ok

Answer 17

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

Answer 18

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

Answer 19

Hint: Simultaneous equations.

Answer 20

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

Answer 21

equate the two and solve for x?

Answer 22

Sorry for taking forever to reply...

Answer 23

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

Answer 24

its alright

Answer 25

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....

Answer 26

we would have to find m first

Answer 27

True. Trying something different...

Answer 28

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

Answer 29

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?

Answer 30

ok

Answer 31

you mean now find the line equations?

Answer 32

Yep.

Answer 33

k

Answer 34

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

Answer 35

doesnt seem right when i graph it...

Answer 36

hopefully u have a better one

Answer 37

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

Answer 38

u graphed mine?

Answer 39

Yep.

Answer 40

it doesnt look tangent to the curve

Answer 41

Plotting on a calculator?

Answer 42

yes

Answer 43

they just all intersect together and dont touch the curve

Answer 44

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...

Answer 45

sure

Answer 46

perhaps i should bump this

Answer 47

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)

Answer 48

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*

Answer 49

i see

Answer 50

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.

Answer 51

alright

Answer 52

ill leave this question open

Answer 53

thanks for helping

Answer 54

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

Answer 55

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

Answer 56

Nice. Good job! :)