f'(1) would be the slope(m).
and the equation of the tangent line can be found by:
In order to find the equation of a tanget line, you need the slope.
the problem I was given is
f(x) = and the point they want me to find it at is (1,-4)
my work so far has been
f'(x) =
f'(x) =
and finally
f'(x) =
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.
however, in the book they gave me the point (1,-4)
so plugging f'(1) I get -3.
and if slope is 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?
The derivate is the slope
ie:
From f(x) I would expand the brackets rather then use the product rule (but both work)
That should be easier to derive
edit: the error was in my derivative
edit2: don't want to make it too easy
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
Now use to find the equation of the tangent line..
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.