Find an equation of the tangent line to
the graph of F at the given point. MEDAL WILL BE REWARDED!
Function: y=x^3+x
Point: (-1,-2)

Question

Find an equation of the tangent line to
the graph of F at the given point. MEDAL WILL BE REWARDED!
Function: y=x^3+x
Point: (-1,-2)

anonymous · Answer

@satellite73

anonymous · Answer

did you find the derivative?

anonymous · Answer

take the derivative $f'(x)$ 
then to find the slope at $-1$ find $m=f'(1)$
then use the point slope formula

anonymous · Answer

oops i meant find $m=f'(-1)$

anonymous · Answer

you good from there?

anonymous · Answer

Im still confused

anonymous · Answer

$$f(x)=x^3+x$$
$$f'(x)=3x^2+1$$
$$f'(-1)=3(-1)^2+1=3+1=4$$your slope is $m=4$

anonymous · Answer

the point is $(-1,-2)$ and so using the point slope formula  your equation for the line is 
$$y+2=4(x+1)$$ or whatever you get when you solve for $y$

anonymous · Answer

So I just plug in the point values into the equation?

anonymous · Answer

not sure what you mean exactly
$f'(x)$ is the formula for the slope
the actual slope is a number, when you get when you compute $f'(-1)$ in this example

anonymous · Answer

Oh, okay, now I see! Thank you so much!!!!

anonymous · Answer

slope is $f'(-1)=4$and point is $(-1,-2)$ and point - slope formula gives equation of the line
yw

anonymous · Answer

Oh okay I completely understand this now

anonymous · Answer

Thank you for making it easier