Question

Find the equations of the tangent line and normal line to the given curve at the specified point.
y = 9xe^x
(0, 0)
Tangent y =
Normal y =

Answer 1

first differentiate y wrt x and sub x = 0 to get the gradient of the tangent and use the standard equation passing thru the origin
repeat the above to get the equation of the normal exc ept that the gradient will be -1 over the gradient of the tangent

Answer 2

y=9x e^x
y'= 9e^x+9x e^x
plug in 0 for x
f'(0)=9
tangent line
y=9x
normal line is perpendicular so
y=-1/9x

Answer 3

Picture contains 9x and -1/9 x (tangent line and normal line)