What is the tangent of (x-2)(x^2-10x+25) ?

The tangent of (x-2)(x^2-10x+25) is y=(3a_{0}^2-24a_{0}+45)x-2a_{0}^3+12a_{0}^2-50

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tangent of (x-2)(x^2-10x+25)

tangent of

(x - 2) (x^{2} - 10 x + 25)

y = (3 a_{0}^{2} - 24 a_{0} + 45) x - 2 a_{0}^{3} + 12 a_{0}^{2} - 50

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (a_{0} - 2) (a_{0}^{2} - 10 a_{0} + 25))

Find the slope of

y = (x - 2) (x^{2} - 10 x + 25) : \frac{d y}{d x} = 3 x^{2} - 24 x + 45

Compute the slope of

y = (x - 2) (x^{2} - 10 x + 25)

x = a_{0} : m = 3 a_{0}^{2} - 24 a_{0} + 45

Find the line with slope m=

3 a_{0}^{2} - 24 a_{0} + 45

and passing through

(a_{0}, (a_{0} - 2) (a_{0}^{2} - 10 a_{0} + 25)) : y = (3 a_{0}^{2} - 24 a_{0} + 45) x - 2 a_{0}^{3} + 12 a_{0}^{2} - 50

y = (3 a_{0}^{2} - 24 a_{0} + 45) x - 2 a_{0}^{3} + 12 a_{0}^{2} - 50

\frac{d x}{d y} = 4 (x^{2} 1), x (\frac{π}{4}) = 1

(3 x^{2} + y - 4) - (2 y - x) \cdot \frac{d y}{d x} = 0

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

\frac{d y}{d x} - \frac{y}{x} = - \frac{5}{2} x^{2} y^{3}

4 (t - 2) e^{4 t}

What is the tangent of (x-2)(x^2-10x+25) ?
The tangent of (x-2)(x^2-10x+25) is y=(3a_{0}^2-24a_{0}+45)x-2a_{0}^3+12a_{0}^2-50