What is the tangent of 5x^3-4x^2+7x-8,\at x=1 ?

The tangent of 5x^3-4x^2+7x-8,\at x=1 is y=14x-14

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tangent of 5x^3-4x^2+7x-8,\at x=1

tangent of

5 x^{3} - 4 x^{2} + 7 x - 8, at x = 1

y = 14 x - 14

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

y = 5 x^{3} - 4 x^{2} + 7 x - 8 : \frac{d y}{d x} = 15 x^{2} - 8 x + 7

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 = 14

Find the line with slope m=

14

and passing through

(1, 0) : y = 14 x - 14

y = 14 x - 14

What is the tangent of 5x^3-4x^2+7x-8,\at x=1 ?
The tangent of 5x^3-4x^2+7x-8,\at x=1 is y=14x-14