Question

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)?

Answer 1

equation of tangent is `y=12x-3`
equation of normal line is `12y = 109-x`

Explanation:

the slope of tangent at (1,9) is `2(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