Question

find an equation of the tangent line at point given: f(x)=8/x^2+4; (2,1)

Answer 1

Have you differentiated f yet?

Answer 2

not sure how

Answer 3

would i have to plug in the point into the point slope form ? or

Answer 4

\[f(x)=\frac{8}{x^2+4} \text{ or } f(x)=\frac{8}{x^2}+4 ?\]

Answer 5

I think you mean the first one because (2,1) isn't on the second graph

Answer 6

To differentiate f you need to know the quotient rule

Answer 7

Or at least the product rule and chain rule

Answer 8

yeah the first one

Answer 9

Do you know the quotient rule?

Answer 10

f'g-fg'/g^2 ?

Answer 11

\[f'(x)=\frac{(8)'(x^2+4)-8(x^2+4)'}{(x^2+4)^2}\]

Answer 12

so you need to know the derivative of 8 and the derivative of x^2+4 to finish that

Answer 13

okie dokie !

Answer 14

once you are done with that replace the x's with 2'
to find f'(2) this will be the slope of the tangent line of f at (2,1)

Answer 15

then once you know the slope that is it plug in the point-slope form of a line
y-y1=m(x-x1)

Answer 16

\[y-1=m(x-2)\]

Answer 17

your only job is to find m
which is f'(2)