Question

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

Answer 1

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

Answer 2

please help me

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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 \(c\)

Answer 10

thank you..

Answer 11

yw