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

The tangent of y=(x+2)(x-1)(x-3) is y=(3a_{0}^2-4a_{0}-5)x-2a_{0}^3+2a_{0}^2+6

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

tangent of

y = (x + 2) (x - 1) (x - 3)

y = (3 a_{0}^{2} - 4 a_{0} - 5) x - 2 a_{0}^{3} + 2 a_{0}^{2} + 6

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (a_{0} + 2) (a_{0} - 1) (a_{0} - 3))

Find the slope of

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

Compute the slope of

y = (x + 2) (x - 1) (x - 3)

x = a_{0} : m = 3 a_{0}^{2} - 4 a_{0} - 5

Find the line with slope m=

3 a_{0}^{2} - 4 a_{0} - 5

and passing through

(a_{0}, (a_{0} + 2) (a_{0} - 1) (a_{0} - 3)) : y = (3 a_{0}^{2} - 4 a_{0} - 5) x - 2 a_{0}^{3} + 2 a_{0}^{2} + 6

y = (3 a_{0}^{2} - 4 a_{0} - 5) x - 2 a_{0}^{3} + 2 a_{0}^{2} + 6

\int t sec^{2} (9 t) d t

\frac{d}{d x} (- 5 x^{2} + 10)

\frac{d}{d x} (\frac{1}{x ^{n}})

y = \frac{4 x ^{2} + 4 x + 8}{x}

\int_{0}^{8} x^{2} d x

What is the tangent of y=(x+2)(x-1)(x-3) ?
The tangent of y=(x+2)(x-1)(x-3) is y=(3a_{0}^2-4a_{0}-5)x-2a_{0}^3+2a_{0}^2+6