What is the tangent of y=x^2+2,(-4,18) ?

The tangent of y=x^2+2,(-4,18) is y=-8x-14

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tangent of y=x^2+2,(-4,18)

tangent of

y = x^{2} + 2, (- 4, 18)

y = - 8 x - 14

Solution steps

Find the slope of

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

Compute the slope of

y = x^{2} + 2

(- 4, 18) : m = - 8

Find the line with slope m=

- 8

and passing through

(- 4, 18) : y = - 8 x - 14

y = - 8 x - 14

y = 7 x, y = \frac{7}{4} x, y = 78 - x^{2}

\frac{\partial}{\partial t} (r \cdot cos (t))

ln (, at^{c})

\int - \frac{1}{50} d x

\frac{d y}{d x} = \frac{1}{x + 1} y + 4 x^{2} + 4 x

What is the tangent of y=x^2+2,(-4,18) ?
The tangent of y=x^2+2,(-4,18) is y=-8x-14