Find equation of tangent line given only x. Please help

[939]

Homework Statement

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.

Homework Equations

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

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?

 

[SammyS]

Homework Statement

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.

Homework Equations

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

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 ?

 

[939]

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?

 

[Mark44]

Homework Statement

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.

Homework 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)

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."

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?

 

[939]

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?

 

[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...

 

[Mark44]

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 = [/color]2x - 33