y=x^2/3 ; (8,4)

Question

y=x^2/3 ; (8,4)

anonymous · Answer

find the equation of the tangent line if it exist to the curve at the given point

anonymous · Answer

please help me

anonymous · Answer

i assume this is 
$$f(x)=x^{\frac{2}{3}}$$ right?

anonymous · Answer

take the derivative, replace $x$ by 8, and that will give you your slope
then use the point slope formula

anonymous · Answer

$$f'(x)=\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3\sqrt[3]{x}}$$
$$f'(8)=\frac{2}{3\sqrt[3]{8}}=\frac{2}{3\times 2}=\frac{1}{3}$$

anonymous · Answer

your slope is $\frac{1}{3}$ your point is $(8,4)$ use the point slope formula to finish 
ok?

anonymous · Answer

ok, but how if we use vertical tangent line.?

anonymous · Answer

i am not sure what you mean by "vertical tangent line"  
the question asks for the equation of the line tangent to the graph
it is not vertical

anonymous · Answer

we know i is not vertical because the slope is $\frac{1}{3}$ a vertical line has not slope and has the equation $x=c$ for some constant $c$

anonymous · Answer

thank you..

anonymous · Answer

yw