y=x^2/3 ; (8,4) - QuestionCove
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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 $$c$$

5 years ago
OpenStudy (anonymous):

thank you..

5 years ago
OpenStudy (anonymous):

yw

5 years ago