What is the tangent of y=(1-x)(x^2-8)^2,(3,-2) ?

The tangent of y=(1-x)(x^2-8)^2,(3,-2) is y=-25x+73

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tangent of y=(1-x)(x^2-8)^2,(3,-2)

tangent of

y = (1 - x) (x^{2} - 8)^{2}, (3, - 2)

y = - 25 x + 73

Solution steps

Find the slope of

y = (1 - x) (x^{2} - 8)^{2} : \frac{d y}{d x} = - 5 x^{4} + 4 x^{3} + 48 x^{2} - 32 x - 64

Compute the slope of

y = (1 - x) (x^{2} - 8)^{2}

(3, - 2) : m = - 25

Find the line with slope m=

- 25

and passing through

(3, - 2) : y = - 25 x + 73

y = - 25 x + 73

\frac{d x}{d t} = - x - x^{2}

ln (3 x + 2)

n \to \infty lim (i = 1 \sum n \frac{i ^{3}}{n ^{4}} + \frac{1}{n})

m (t) = 5 t (3 t^{4} - 1)^{5}

\frac{d}{d x} (x cos (y) - y e^{x})

What is the tangent of y=(1-x)(x^2-8)^2,(3,-2) ?
The tangent of y=(1-x)(x^2-8)^2,(3,-2) is y=-25x+73