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

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• Oct 10th 2010, 10:59 AM
Chimera
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.
• Oct 10th 2010, 11:07 AM
harish21
f'(1) would be the slope(m).

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

$\displaystyle (y-y_1) = m(x-x_1)$
• Oct 10th 2010, 11:11 AM
Chimera
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?
• Oct 10th 2010, 11:16 AM
e^(i*pi)
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
• Oct 10th 2010, 11:19 AM
harish21
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..
• Oct 10th 2010, 11:49 AM
Chimera
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.
• Oct 10th 2010, 11:51 AM
harish21
f'(1)=-3. That is all you needed to do to find the slope!