What is the equation of the line tangent to y=(x+3)^1/3 at x=-3? How do I find this?
I am struggling. Thank you!

Question

anonymous · Answer

dy/dx = ((x+3)^(-2/3))/3 
therefore at x = -3
-3+3 = 0
= 0^n = 0 
hence 1/0 = ND

anonymous · Answer

This is odd. However it should make sense considering y = 0 when x = -3...

anonymous · Answer

Ok. I find the critical points (0,-3) earlier but is there an equation? I've been doing these all morning and I'm thoroughly confused haha.

hartnn · Answer

so you got dy/dx at x=-3 as 0, right ?

hartnn · Answer

that means your slope of the line = m = 0

hartnn · Answer

you just need y- intercept 
when x= -3
what is the value of y ?

anonymous · Answer

@hartnn 0?...

hartnn · Answer

thats correct!
so,
in y = mx+c 
m=0, c=0
so, whats the equation ?

anonymous · Answer

@hartnn Do I plug those in to y=mx+c? y=0(x)+0

hartnn · Answer

yes

hartnn · Answer

y=0 
x axis
thats it! thats your equation of line :)

anonymous · Answer

Okay, thank you both SO much