Question

Find the equation of the tangent line and normal line to the curve y= (6+3x)^2 at the point (1,81). Tangent line: y=
Normal line: y=

Answer 1

y'=2(6+3x)*3=18x+36
y'(1)=18+36=54=slope

Answer 2

m=(y-y_1)/(x-x_1)
54=(y-81)/(x-1)
find y..

Answer 3

Do you need more help?

Answer 4

Yes Please!

Answer 5

Do you understand the slope value 54?

Answer 6

Yes (:

Answer 7

When you want to find tangent => find first derivative!

Answer 8

Right

Answer 9

Value of slope => constant number of derivative at tangent point!

Answer 10

Do you follow me?

Answer 11

I do yes!

Answer 12

Can you find derivative for me?

Answer 13

18(x+2)

Answer 14

Tell me where you have the trouble?

Answer 15

Yep, you're right :)
f'(x) = 18 x + 36

Answer 16

Since tangent line is linear which y = mx + b
=> We need slope m, and the tangent point!

Answer 17

Can you find slop m?

Answer 18

Not sure how to

Answer 19

Just plug the value of tangent point in:
At ( 1, 81),
f'(1) = ...

Answer 20

54

Answer 21

Yep f'(1) = 18 * 1 + 36 = 54
Are you sure you understand how to find slope now?

Answer 22

Yes thank you for your help