1. The problem statement, all variables and given/known data

Find equation of tangent line, given x = -1. Not given y. I am used to having this when I am given both y and x.

2. Relevant equations

(x^3 - 4x + 2)(x^4 + 3x - 5)

3. The attempt at a solution

Differentiate
(3x^2 - 4)(4x^3+3)

Multiply
12x^5 - 9x^2 - 8x^3 - 12

Plug in -1, find slope
12(-1)^5 - 9 (-1)^2 - 8(-1)^3 - 12
= -5.3

Plug in -1 to the original equation to find Y

-35

Find equation
y1- = m (x1 - x2)
y+35 = -5.3 (x+1)
y+35 = -5.3x -5.3
y = -5.3x - 5.3 - 35
y = -5.3x -40.3

...............................

I have a feeling this is wrong though, because I have gotten many similar questions wrong. How can I figure this out?
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity

Mentor
 Quote by 939 1. The problem statement, all variables and given/known data Find equation of tangent line, given x = -1. Not given y. I am used to having this when I am given both y and x. 2. Relevant equations (x^3 - 4x + 2)(x^4 + 3x - 5) 3. The attempt at a solution Differentiate (3x^2 - 4)(4x^3+3) Multiply 12x^5 - 9x^2 - 8x^3 - 12 Plug in -1, find slope 12(-1)^5 - 9 (-1)^2 - 8(-1)^3 - 12 = -5.3 Plug in -1 to the original equation to find Y -35 Find equation y1- = m (x1 - x2) y+35 = -5.3 (x+1) y+35 = -5.3x -5.3 y = -5.3x - 5.3 - 35 y = -5.3x -40.3
Hello 939. Welcome to PF !

When you graph the function, (x3 - 4x + 2)(x4 + 3x - 5) , what is y ?

 Quote by SammyS Hello 939. Welcome to PF ! When you graph the function, (x3 - 4x + 2)(x4 + 3x - 5) , what is y ?
Hi Sammy, thanks!

When I graph it, y is -35 exactly.

Knowing y and x, is it possible to find the equation doing:

y-y1 = m (x-x1)
y+35 = -5.3 (x+1)
etc?

Mentor

 Quote by 939 1. The problem statement, all variables and given/known data Find equation of tangent line, given x = -1. Not given y. I am used to having this when I am given both y and x. 2. Relevant equations (x^3 - 4x + 2)(x^4 + 3x - 5)
I'm assuming that the equation is y = (x^3 - 4x + 2)(x^4 + 3x - 5)

 Quote by 939 3. The attempt at a solution Differentiate (3x^2 - 4)(4x^3+3)
If y = f(x) * g(x), dy/dx $\neq$ f'(x) * g'(x), which is what you seem to be doing here. You should be thinking "product rule."
 Quote by 939 Multiply 12x^5 - 9x^2 - 8x^3 - 12 Plug in -1, find slope 12(-1)^5 - 9 (-1)^2 - 8(-1)^3 - 12 = -5.3 Plug in -1 to the original equation to find Y -35 Find equation y1- = m (x1 - x2) y+35 = -5.3 (x+1) y+35 = -5.3x -5.3 y = -5.3x - 5.3 - 35 y = -5.3x -40.3 ............................... I have a feeling this is wrong though, because I have gotten many similar questions wrong. How can I figure this out?

 Quote by Mark44 I'm assuming that the equation is y = (x^3 - 4x + 2)(x^4 + 3x - 5) If y = f(x) * g(x), dy/dx $\neq$ f'(x) * g'(x), which is what you seem to be doing here. You should be thinking "product rule."
Sorry, yes you are right about the equation. And regarding the product rule yes, you are also right!

So, when I find it using the product rule, I insert the -1, and then that gives me the slope?
 So the product and then plugging in gives me 2... Provided I did that right, can I find the equation by doing: y+35 = 2(x+1) = 2x+2 - 35, I believe...

Mentor
 Quote by 939 So the product and then plugging in gives me 2... Provided I did that right, can I find the equation by doing: y+35 = 2(x+1) = 2x+2 - 35, I believe...
Your arithmetic is correct, but you're being sloppy in how you present it.

Keep your equations separated. The first equation is y+35 = 2(x+1)
Add -35 to both sides to get a new equation y = 2x - 33

 Tags calculus, equation, slope, tangent line

 Similar discussions for: Find equation of tangent line given only x. Please help! Thread Forum Replies Calculus & Beyond Homework 6 Calculus & Beyond Homework 1 Calculus & Beyond Homework 1 Calculus & Beyond Homework 3 Calculus & Beyond Homework 3