What is the tangent of f(x)=x^4-8x^3+4,\at x=-1 ?

The tangent of f(x)=x^4-8x^3+4,\at x=-1 is y=-28x-15

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tangent of f(x)=x^4-8x^3+4,\at x=-1

tangent of

f (x) = x^{4} - 8 x^{3} + 4, at x = - 1

y = - 28 x - 15

Solution steps

Find the tangent point:

(- 1, 13)

Find the slope of

f (x) = x^{4} - 8 x^{3} + 4 : \frac{df ( x )}{d x} = 4 x^{3} - 24 x^{2}

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 = - 28

Find the line with slope m=

- 28

and passing through

(- 1, 13) : y = - 28 x - 15

y = - 28 x - 15

What is the tangent of f(x)=x^4-8x^3+4,\at x=-1 ?
The tangent of f(x)=x^4-8x^3+4,\at x=-1 is y=-28x-15