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):

please help me

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