What is the tangent of y=(x^3-2)(x^2-3x+1),(2,-6) ?

The tangent of y=(x^3-2)(x^2-3x+1),(2,-6) is y=-6x+6

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tangent of y=(x^3-2)(x^2-3x+1),(2,-6)

tangent of

y = (x^{3} - 2) (x^{2} - 3 x + 1), (2, - 6)

y = - 6 x + 6

Solution steps

Find the slope of

y = (x^{3} - 2) (x^{2} - 3 x + 1) : \frac{d y}{d x} = 5 x^{4} - 12 x^{3} + 3 x^{2} - 4 x + 6

Compute the slope of

y = (x^{3} - 2) (x^{2} - 3 x + 1)

(2, - 6) : m = - 6

Find the line with slope m=

- 6

and passing through

(2, - 6) : y = - 6 x + 6

y = - 6 x + 6

\frac{d x}{d t} = 7 x

4 x + 3 x^{2}

(1 - x^{2})^{- \frac{1}{2}}

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

\int \frac{x - 17}{x ^{2} + x - 12} d x

What is the tangent of y=(x^3-2)(x^2-3x+1),(2,-6) ?
The tangent of y=(x^3-2)(x^2-3x+1),(2,-6) is y=-6x+6