What is the tangent of (x^2-60)(2x-3) ?

The tangent of (x^2-60)(2x-3) is y=(6a_{0}^2-6a_{0}-120)x-4a_{0}^3+3a_{0}^2+180

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tangent of (x^2-60)(2x-3)

tangent of

(x^{2} - 60) (2 x - 3)

y = (6 a_{0}^{2} - 6 a_{0} - 120) x - 4 a_{0}^{3} + 3 a_{0}^{2} + 180

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (a_{0}^{2} - 60) (2 a_{0} - 3))

Find the slope of

y = (x^{2} - 60) (2 x - 3) : \frac{d y}{d x} = 6 x^{2} - 6 x - 120

Compute the slope of

y = (x^{2} - 60) (2 x - 3)

x = a_{0} : m = 6 a_{0}^{2} - 6 a_{0} - 120

Find the line with slope m=

6 a_{0}^{2} - 6 a_{0} - 120

and passing through

(a_{0}, (a_{0}^{2} - 60) (2 a_{0} - 3)) : y = (6 a_{0}^{2} - 6 a_{0} - 120) x - 4 a_{0}^{3} + 3 a_{0}^{2} + 180

y = (6 a_{0}^{2} - 6 a_{0} - 120) x - 4 a_{0}^{3} + 3 a_{0}^{2} + 180

\int \frac{x - 4}{x + 2} d x

\int (2 t + 1) d t

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

81 y^{^{''}} + 72 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

\int_{1}^{2} (\frac{3}{x ^{2}} - 1) d x

What is the tangent of (x^2-60)(2x-3) ?
The tangent of (x^2-60)(2x-3) is y=(6a_{0}^2-6a_{0}-120)x-4a_{0}^3+3a_{0}^2+180