Question

find the equation of the line tangent to y=(3x/2+2)^2 at x=2.

Answer 1

dy/dx=2*(3/2)*(3x/2+2)
at x=2
dy/dx=15
at x=2 y=25

so y-25=15(x-2)

Answer 2

what's the first step you did?

Answer 3

find the derivative of y=(3x/2+2)^2

Answer 4

ok

Answer 5

got it??

Answer 6

yes, thank you very much :) thanks to this site i will be able to get the subjects i dont get..i have my final on friday and i need to pass it

Answer 7

you most welcome.......

Answer 8

is the answer in this form (y=mx+b) this ... y=15x+20 ??

Answer 9

no..
y-25=15x-30
y=15x-5

Answer 10

why is it "-5" instead of "+20" ?

Answer 11

how did you get y=15x+20..can u please show me..my calculation says its y=15x-5

Answer 12

i solved for "b" ....after getting y'=15(which is what you got) i set equal to zero this..y'=3(3x/2+2) and got x=-4/3..i forgot how i got zero for "y" and then plugged in x and y and the slope and got b=20

Answer 13

why u set equal to zero this..y'=3(3x/2+2) ??

Answer 14

so i could get "x"