What is the tangent of f(x)=(x-4)(x^2+8),(1,-27) ?

The tangent of f(x)=(x-4)(x^2+8),(1,-27) is y=3x-30

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tangent of f(x)=(x-4)(x^2+8),(1,-27)

tangent of

f (x) = (x - 4) (x^{2} + 8), (1, - 27)

y = 3 x - 30

Solution steps

Find the slope of

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

Compute the slope of

f (x) = (x - 4) (x^{2} + 8)

(1, - 27) : m = 3

Find the line with slope m=

3

and passing through

(1, - 27) : y = 3 x - 30

y = 3 x - 30

x \to 1 - lim (- x^{2} + 5 x + (- 4))

n = 3 \sum \infty \frac{2}{n ln ^{2} ( n )}

y = x, y = 3 x

\int \frac{x}{( 5 x ^{2} + 2 ) ^{6}} d x

\int_{0}^{1} \frac{1}{1 + x} d x

What is the tangent of f(x)=(x-4)(x^2+8),(1,-27) ?
The tangent of f(x)=(x-4)(x^2+8),(1,-27) is y=3x-30