Question

Find the value of x at which the tangent line to the curve y=1/(x+8) passes through the origin.

Answer 1

say we find the tangent line at x=a
\[y-f(a)=f'(a)(x-a) \\ y=f'(a)(x-a)+f(a)\]
can you find the y-intercept of this tangent line?

Answer 2

I'm sorry. I am really confused :(

Answer 3

do you know where I got y=f'(a)(x-a)+f(a) ?

Answer 4

the point slope form of a line is
y-f(a)=m(x-a) agree?
if we know this line is tangent to y=1/(x+8) at x=a
then m=f'(a)

y-f(a)=f'(a)(x-a)
adding f(a) on both sides gives
y=f'(a)(x-a)+f(a)

Answer 5

now actually we are given this line goes through point (0,0)
so just enter in (0,0) for (x,y)

Answer 6

oh

Answer 7

and you will get an equation just in terms of a

Answer 8

then you solve that equation for a

Answer 9

what about f'(a), do i plug in 0?

Answer 10

no

Answer 11

f'(a) is the slope at x=a

Answer 12

you are given f(x)=1/(x+8)
you must find f'(x)

Answer 13

f'(x)=-1/(x+8)^2

Answer 14

right and that evaluated at a is -1/(a+8)^2

Answer 15

right

Answer 16

\[y=\frac{-1}{(a+8)^2}(x-a)+f(a)\]
you still know this line goes through (0,0) so you still need to enter in (0,0) for (x,y)
and I bet you know that f(a)=1/(a+8)

Answer 17

\[0=\frac{-1}{(a+8)^2}(0-a)+\frac{1}{a+8}\]
solve for a

Answer 18

a=-4?

Answer 19

looks good

Answer 20

that makes so much more sense now :D, thank you!

Answer 21

you can check your work by pluggin back -4 for a into y=f'(a)(x-a)+f(a)
\[y=\frac{-1}{(-4+8)^2}(x-(-4))+\frac{1}{-4+8} \\ y=\frac{-1}{4^2}(x+4)+\frac{1}{4} \\ y=\frac{-1}{4^2}x-\frac{1}{4}+\frac{1}{4} \\ y=\frac{-1}{4^2}x\]
yep this line definitely passes through (0,0)

Answer 22

y=mx+b
if b is 0 then (0,0) is on the line y=mx+b