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

The tangent of y=(1-x)(x^2-3)^2,(2,-1) is y=-9x+17

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

tangent of

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

y = - 9 x + 17

Solution steps

Find the slope of

y = (1 - x) (x^{2} - 3)^{2} : \frac{d y}{d x} = - 5 x^{4} + 4 x^{3} + 18 x^{2} - 12 x - 9

Compute the slope of

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

(2, - 1) : m = - 9

Find the line with slope m=

- 9

and passing through

(2, - 1) : y = - 9 x + 17

y = - 9 x + 17

\frac{1}{1 - x}, 19

\int x^{2} lo g_{3} (x) d x

\int x^{2} + e^{x} d x

y = \frac{x}{8 x - 4}, at x = 2

x^{4} - 2

What is the tangent of y=(1-x)(x^2-3)^2,(2,-1) ?
The tangent of y=(1-x)(x^2-3)^2,(2,-1) is y=-9x+17