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

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Answer 1 · 2017-03-07T16:53:55

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

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

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

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

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

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

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

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

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

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