Help with example of finding equation of tangent line

[mdem1234]

I'm reading an example to try and figure out how to find the equation of the tangent line, the example goes like this:

Find the equation of the tangent line to f(x) = x^3 - 4x + 1 at point x=1

The example says we calculate the derivative which i understand..

f'(x) = 3x^2 - 4

and then subbing the point -1 into the equation to get the gradient, which i also understand..

f'(-1)= 3(1)^2 - 4 = 3-4 = -1

Then it says we sub 1 into the original equation because we know the tangent line touches the graph at x=1, and by doing this we get the point (1,-2)

Then it says to find the equation, but this is where Im stuck.. The example does it like this but I dont understand...

y-(-2)/x-1 = -1

y+2=1-x

y=-x-1


can anyone explain how doing this finds the equation? And is there another way to find the equation?
much appreciated

 

[agilder23]

Well, by finding the derivative of the equation and plugging in x=1 you get the point -1.

f'(x)=3x^2-4
f'(1)=3(1)^2-4
f'(1)=3-4
f'(1)=-1
m=-1
Note: The -1 just found is the slope of the tangent line.

Next, you take the -2 from the point (1,-2) that you found by subbing 1 in the original equation and the -1 that we just found and enter it into the equation of a line:
y-y1= m (x-x1)
y-(-2)= (-1)(x-1)

So you get:
y+2=-1(x-1)
Distribute the -1 to (x-1)
y+2=-x+1
Subtract 2 from both sides and you get:
y=-x-1



Hope this helps some!

 

[royhaas]

The tangent line is given in point-slope form: \(\displaystyle y = y_0 + m(x - x_0)\), where \(\displaystyle m = f'(x_0), y_0 = f(x_0)\).

 

[mdem1234]

Thank you!