m1cadc

2022-02-10

How do you find the equation of the tangent line and normal line to the curve at the given point $y={(1+2x)}^{2}$ , (1,9)?

Ethen Welch

Beginner2022-02-11Added 11 answers

Explanation:

the slope of tangent at (1,9) is (1+2x)*2 by normal differentiation

at x=1 the slope value is 12

let the equation of tangent is y=mx+c

substituiting points

9=12+c

c=-3

equation of tangent is y=12x-3

for normal line

slope =-1/12

from the line formula y=mx+c

substituiting points

9=-1/12+c

c=109/12

equation is

12y = 109-x

the slope of tangent at (1,9) is (1+2x)*2 by normal differentiation

at x=1 the slope value is 12

let the equation of tangent is y=mx+c

substituiting points

9=12+c

c=-3

equation of tangent is y=12x-3

for normal line

slope =-1/12

from the line formula y=mx+c

substituiting points

9=-1/12+c

c=109/12

equation is

12y = 109-x

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