What is the tangent of f(x)=x^3-5x^2,\at x=2 ?

The tangent of f(x)=x^3-5x^2,\at x=2 is y=-8x+4

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tangent of f(x)=x^3-5x^2,\at x=2

tangent of

f (x) = x^{3} - 5 x^{2}, at x = 2

y = - 8 x + 4

Solution steps

Find the tangent point:

(2, - 12)

Find the slope of

f (x) = x^{3} - 5 x^{2} : \frac{df ( x )}{d x} = 3 x^{2} - 10 x

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 = - 8

Find the line with slope m=

- 8

and passing through

(2, - 12) : y = - 8 x + 4

y = - 8 x + 4

What is the tangent of f(x)=x^3-5x^2,\at x=2 ?
The tangent of f(x)=x^3-5x^2,\at x=2 is y=-8x+4