What is the tangent of y= 1/2 x^2,(-1,-1) ?

The tangent of y= 1/2 x^2,(-1,-1) is y=(sqrt(3)-1)x+sqrt(3)-2 \lor y=(-sqrt(3)-1)x-2-sqrt(3)

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tangent of y= 1/2 x^2,(-1,-1)

tangent of

y = \frac{1}{2} x^{2}, (- 1, - 1)

y = (3 - 1) x + 3 - 2 or y = (- 3 - 1) x - 2 - 3

Solution steps

Slope of the tangent line at

(- 1, - 1) : m = \frac{x ^{2} + 2}{2 ( x + 1 )}

Find the slope of

y = \frac{1}{2} x^{2} : m = x

Solve

x = \frac{x ^{2} + 2}{2 ( x + 1 )} : x = - 1 + 3, x = - 1 - 3

Find the tangent line at

x = - 1 + 3 : y = (3 - 1) x + 3 - 2

Find the tangent line at

x = - 1 - 3 : y = (- 3 - 1) x - 2 - 3

y = (3 - 1) x + 3 - 2 or y = (- 3 - 1) x - 2 - 3

\frac{d y}{d t} + 3 y = 1

\int_{0}^{2} 4 x^{7} d x

d er i v a t i v e 7^{9 x}

d er i v a t i v e 5 r

\int (\frac{4}{cos ^{2} ( x )} - \frac{5}{sin ^{2} ( x )}) d x

What is the tangent of y= 1/2 x^2,(-1,-1) ?
The tangent of y= 1/2 x^2,(-1,-1) is y=(sqrt(3)-1)x+sqrt(3)-2 \lor y=(-sqrt(3)-1)x-2-sqrt(3)