# 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)?

Sep 16, 2015

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

#### Explanation:

the slope of tangent at (1,9) is $2 \left(1 + 2 x\right) \cdot 2$ 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