What is the tangent of P=x^3-5x,(1,-4) ?

The tangent of P=x^3-5x,(1,-4) is y=-2x-2

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tangent of P=x^3-5x,(1,-4)

tangent of

P (x) = x^{3} - 5 x, (1, - 4)

y = - 2 x - 2

Solution steps

Find the slope of

P (x) = x^{3} - 5 x : \frac{d P ( x )}{d x} = 3 x^{2} - 5

Compute the slope of

P (x) = x^{3} - 5 x

(1, - 4) : m = - 2

Find the line with slope m=

- 2

and passing through

(1, - 4) : y = - 2 x - 2

y = - 2 x - 2

cos (\frac{1 - e ^{8 x}}{1 + e ^{8 x}})

f (x) = (x^{2} + 3 x + 5) e^{6 x}

\frac{d}{d t} (2 - 2 t)

\int \frac{csc ^{2} ( x )}{cot ^{2} ( x )} d x

\int_{0}^{2} \frac{e ^{x}}{1 + e ^{x}} d x

What is the tangent of P=x^3-5x,(1,-4) ?
The tangent of P=x^3-5x,(1,-4) is y=-2x-2