anonymous
  • anonymous
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=
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
y'=2(6+3x)*3=18x+36 y'(1)=18+36=54=slope
anonymous
  • anonymous
m=(y-y_1)/(x-x_1) 54=(y-81)/(x-1) find y..
anonymous
  • anonymous
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anonymous
  • anonymous
Yes Please!
anonymous
  • anonymous
Do you understand the slope value 54?
anonymous
  • anonymous
Yes (:
anonymous
  • anonymous
When you want to find tangent => find first derivative!
anonymous
  • anonymous
Right
anonymous
  • anonymous
Value of slope => constant number of derivative at tangent point!
anonymous
  • anonymous
Do you follow me?
anonymous
  • anonymous
I do yes!
anonymous
  • anonymous
Can you find derivative for me?
anonymous
  • anonymous
18(x+2)
anonymous
  • anonymous
Tell me where you have the trouble?
anonymous
  • anonymous
Yep, you're right :) f'(x) = 18 x + 36
anonymous
  • anonymous
Since tangent line is linear which y = mx + b => We need slope m, and the tangent point!
anonymous
  • anonymous
Can you find slop m?
anonymous
  • anonymous
Not sure how to
anonymous
  • anonymous
Just plug the value of tangent point in: At ( 1, 81), f'(1) = ...
anonymous
  • anonymous
54
anonymous
  • anonymous
Yep f'(1) = 18 * 1 + 36 = 54 Are you sure you understand how to find slope now?
anonymous
  • anonymous
Yes thank you for your help

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