What is the tangent of y=-x^2,(8,-64) ?

The tangent of y=-x^2,(8,-64) is y=-16x+64

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tangent of y=-x^2,(8,-64)

tangent of

y = - x^{2}, (8, - 64)

y = - 16 x + 64

Solution steps

Find the slope of

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

Compute the slope of

y = - x^{2}

(8, - 64) : m = - 16

Find the line with slope m=

- 16

and passing through

(8, - 64) : y = - 16 x + 64

y = - 16 x + 64

e^{x}, e^{- 4 x}, ln (4)

\int - x^{2} + x + 6 d x

y = x + 3, x = - 3, y = 5

\frac{\partial}{\partial x} (2000 e^{(x) (y)})

t^{2} + a^{2}

What is the tangent of y=-x^2,(8,-64) ?
The tangent of y=-x^2,(8,-64) is y=-16x+64