1. Feb 12, 2012

### 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

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

I have a feeling this is wrong though, because I have gotten many similar questions wrong. How can I figure this out?

2. Feb 12, 2012

### SammyS

Staff Emeritus
Hello 939. Welcome to PF !

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

3. Feb 12, 2012

### 939

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?

Last edited: Feb 13, 2012
4. Feb 13, 2012

### Staff: Mentor

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."

5. Feb 13, 2012

### 939

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?

6. Feb 13, 2012

### 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...

Last edited: Feb 13, 2012
7. Feb 13, 2012

### Staff: Mentor

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