Question

What is the equation of the line that is normal to `f(x)=-2x^2-2e^x` at `x=-2`?

Answer 1

`f(x)=-2x^2-2e^x`

`f'(x)=-4x-2e^x`

`f'(-2)=8-2/e^2 larr` This is the gradient of the tangent.

`"m"_"n"=-1/(m_tan)=-1/(8-2/e^2)=-(e^2)/(8e^2-2)`

`f(-2)=-8-2/e^2=-(8e^2-2)/e^2`

`y-y_1=m(x-x_1)`

`y+(8e^2-2)/e^2=-(e^2)/(8e^2-2)(x+2)`

`y+(8e^2-2)/e^2=-(xe^2)/(8e^2-2)-(2e^2)/(8e^2-2)`

`y=(xe^2)/(1-8e^2) -(33e^4 -16e^2+2)/(e^2(2e-1)(2e+1))`