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

The tangent of f(x)=(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 f(x)=(x-2)(x^2-10x+25)

tangent of

f (x) = (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

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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

e^{t} sin (t)

f (t) = t^{2} + 1

n = 8 \sum \infty e^{5 - 6 n}

\int_{0}^{\frac{3}{2}} cos (π x) d x

y^{^{'}} = e^{x}

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