What is the tangent of-2x^2,\at x=-2 ?

The tangent of-2x^2,\at x=-2 is y=8x+8

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tangent of-2x^2,\at x=-2

tangent of

- 2 x^{2}, at x = - 2

y = 8 x + 8

Solution steps

Find the tangent point:

(- 2, - 8)

Find the slope of

y = - 2 x^{2} : \frac{d y}{d x} = - 4 x

Compute the slope of

y = - 2 x^{2}

x = - 2 : m = 8

Find the line with slope m=

8

and passing through

(- 2, - 8) : y = 8 x + 8

y = 8 x + 8

\frac{1}{x ^{2}}

\frac{12}{s ^{2} - 5}

1 - 4 x

y^{^{'}} = \frac{y}{x} + \frac{y ^{2}}{x ^{2}}

- \frac{1}{2} y^{^{'}} + y = 4