# Thread: Finding the equation of a tanget line(derivative error)

1. ## Finding the equation of a tanget line(derivative error)

In order to find the equation of a tanget line, you need the slope.

the problem I was given is

f(x) = $\displaystyle ((x^3)+4x-1)(x-2)$ and the point they want me to find it at is (1,-4)
my work so far has been
f'(x) =$\displaystyle (x-2)((3x^2)+4) + ((x^3) + 4x-1)$
f'(x) =$\displaystyle ((3x^3)+4x-(6x^2)-8 + (x^3) +4x -1$
and finally
f'(x) =$\displaystyle (4x^3)+8x-(6x^2)-9$
Would I be correct in plugging two points into my final f'(x) in order to get the slope?

I thought I was done but, something is off in my logic and I'm not sure what.

2. f'(1) would be the slope(m).

and the equation of the tangent line can be found by:

$\displaystyle (y-y_1) = m(x-x_1)$

3. however, in the book they gave me the point (1,-4)

so plugging f'(1) I get -3.

and if slope is $\displaystyle rise/run$ then doesn't it make sense I should have to plug something else in as well? I tried however, I have not gotten the answer in the back of the book which is y = -3x -1..

So I am wondering if I perhaps did my derivative wrong?

4. The derivate is the slope

ie: $\displaystyle \dfrac{\text{ rise}}{\text{ run}} = f^{\prime}(x)$

From f(x) I would expand the brackets rather then use the product rule (but both work)

$\displaystyle (x-2)(x^3+4x-1) = (x^4+4x^2-x) + (-2x^3-8x+2) = x^4-2x^3+4x^2-9x+2$

That should be easier to derive

$\displaystyle f'(x) = 4x^3-6x^2+8x-9$

$\displaystyle f'(1) = 4-6+8-9 = -3$

edit: the error was in my derivative
edit2: don't want to make it too easy

5. Why doesn't it make sense? Do you mean that the slope of a line cannot be negative??

When a line has a positive slope it goes up left to right.

When a line has a negative slope it goes down left to right.

You have mentioned the answer at the back of the book is y = -3x-1..

So your slope is m = -3.

The point you have been given is $\displaystyle (x_{1}, y_{1}) = (1,-4)$

Now use $\displaystyle (y-y_1) = m(x-x_1)$ to find the equation of the tangent line..

6. So wait, if I have the derivative (which seems to be correct).

and I am given the point (1,-4) and I plug 1 into the derivative and get (1, -3), that is all I need to do for slope?

oh x.x duh.. -3 is the slope at 1,-4... now it makes a bit more sense.. lol. totally was not thinking about the fact that by plugging 1 into the derivative it was giving me the slope.. I for some reason was thinking I would have to plug another number like 2 into it and get the answer.. but, I'm mistaken.. laughable so.

7. f'(1)=-3. That is all you needed to do to find the slope!