Question

Find an equation for the tangent line to the given curve at the point where x=x[0].
Give your answer in Y=mx+b form, where m and b are simplified fractions.
y=(x)/(7x+9) ; x[0] =-1


Please help is calculus!!

Answer 1

hi did you get the derivative of the equation?
y=(x)/(7x+9)

Answer 2

vdu/dx-udv/dx
use y' = ---------------
v^2

Answer 3

How do i use it. I am stuck right at this place.
(7)/(7x+9)^(2)

Answer 4

ok that was y'=m=(9)/(7x+9)^(2)]x=-1 solve this one with x=-1 and that will be your m=slope

Answer 5

Thx, I can get thru that part. But they want the tangent line. In fractions thats why I am confused.

Answer 6

your m=slope=9/4 then use this m on y-y1=m(x-x1)

Answer 7

Is x[0]=-1

Answer 8

ok at x=x(0)=-1 your y(-1)=(-1)/(7(-1)+9) =-1/2.. now youre at position (x1,y1)=(-1,-1/2) use this also including you value of m=9/4, at the eq of line y-y1=m(x-x1)

Answer 9

now your eq of the line tangent will be
y-y1=m(x-x1)
y-(-1/2)=(9/4)(x-(-1))
y+1=(9/4)(x+1) eq of the tangent line

Answer 10

Thx, Not sure why they got 2 1/4x + 1 3/4 as an answer. I am lost.

Answer 11

or you can put it in the form of Y=mx+b form,

Answer 12

Y=mx+b
y+1=(9/4)(x+1)
y=(9/4)(x+1) +1 now solve for y
y=9x/4 +9/4 +1
y=9x/4 +13/4 ans

Answer 13

Thanks so much, i will copy this steps, but the answer was different.

Answer 14

ok yw,, gud luck now

Answer 15

I ll keep trying! thx again..