What is the tangent of y=(x-4)(x^2+8),(1,-27) ?

The tangent of y=(x-4)(x^2+8),(1,-27) is y=3x-30

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tangent of y=(x-4)(x^2+8),(1,-27)

tangent of

y = (x - 4) (x^{2} + 8), (1, - 27)

y = 3 x - 30

Solution steps

Find the slope of

y = (x - 4) (x^{2} + 8) : \frac{d y}{d x} = 3 x^{2} - 8 x + 8

Compute the slope of

y = (x - 4) (x^{2} + 8)

(1, - 27) : m = 3

Find the line with slope m=

3

and passing through

(1, - 27) : y = 3 x - 30

y = 3 x - 30

\frac{d}{d x} (e^{x} - 1 + e^{x} x)

\frac{\partial}{\partial x} (x^{0.2} y^{0.8})

\int (4 x + 3) (2 x^{2} + 3)^{3} d x

\frac{\partial}{\partial z} (x^{2} y^{3} z)

\frac{d}{d x} (\frac{1}{2 + x})

What is the tangent of y=(x-4)(x^2+8),(1,-27) ?
The tangent of y=(x-4)(x^2+8),(1,-27) is y=3x-30