What is the tangent of y=4-x^2,(4,-12) ?

The tangent of y=4-x^2,(4,-12) is y=-8x+20

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tangent of y=4-x^2,(4,-12)

tangent of

y = 4 - x^{2}, (4, - 12)

y = - 8 x + 20

Solution steps

Find the slope of

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

Compute the slope of

y = 4 - x^{2}

(4, - 12) : m = - 8

Find the line with slope m=

- 8

and passing through

(4, - 12) : y = - 8 x + 20

y = - 8 x + 20

\int \frac{6 x}{( 9 x ^{2} - 4 ) ^{3}} d x

\frac{\partial}{\partial x} x y \cdot e^{(\frac{( - 5 x ^{2} + 5 y ^{2} )}{2} + 1)}

\frac{d y}{d x} = y^{2} (6 - 2 x)

\frac{\partial}{\partial x} (ln (\frac{x}{k}))

\frac{d}{d x} ((2 - x) (y - 2) (3 - x - y))

What is the tangent of y=4-x^2,(4,-12) ?
The tangent of y=4-x^2,(4,-12) is y=-8x+20