Question

equation for the line?

Answer 1

point slope form: y - y1 = m(x - x1)
when m = slope and given the point (x1, y1)

slope-intercept form: y = mx + b
when m = slope and b = y-intercept

Answer 2

\[y-2=\sec(\pi/3)\tan(\pi/3)[(x-(\pi/3)]\]

Answer 3

ehhhh was that what you were given?

Answer 4

I am trying to find the equation of the tan line y=secx , (pi/3 , 20

Answer 5

(pi/3 , 2)

Answer 6

equation of a line tangent to y = secx ?
so calculus?

Answer 7

ye

Answer 8

ok :)
when they say "tangent" they do Not mean "tan(x)" ;P

then do you know what the derivative of secx is?

Answer 9

yes sexxtanx

Answer 10

secxtanx

Answer 11

lol yeah ^_^
so f'(x) = secxtanx
then what is f'(pi/3) = ?

Answer 12

when x equals (pi/3)

Answer 13

mhmm :)

Answer 14

oh got it

Answer 15

yep, that would become your slope
which you can then plug into the point-slope form :)

Answer 16

tan is sin/cos right?

Answer 17

yeah, just think of the 30 60 90 triangle

Answer 18

thanks why you so smart!

Answer 19

can I add you on somewhere for cal help?

Answer 20

mmm my email is jigglypuff314@gmail.com
if you need me :P

I only know all this calc cuz I took AP Calc AB last year ^_^"

Answer 21

you got it jugglepuff :)

Answer 22

well, tag me if you need help @Babybunch see ya around ^_^