The idea is to use the tangent line to
at
in order to approximate
.
We have


so the linear approximation to
is

Hence
and
.
Then

Answer:
Step-by-step explanation:
(0, -9) and (3,3)
(3+9)/(3-0) = 12/3 = 4
y + 9 = 4(x - 0)
y + 9 = 4x - 0
y = 4x - 9
Uhm, I'm pretty new to expanded form but I think it's something like:
(3 × 1/10) + (1 × 1/100) + (6 × 1/1000)
You can also write it as:
0.3 + 0.01 + 0.006
Let me know if you need working or anything!
1.7(j-2)
Hope it helped :)