What is the tangent of f(x)=(x-2)(x^2+4),(1,-5) ?

The tangent of f(x)=(x-2)(x^2+4),(1,-5) is y=3x-8

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tangent of f(x)=(x-2)(x^2+4),(1,-5)

tangent of

f (x) = (x - 2) (x^{2} + 4), (1, - 5)

y = 3 x - 8

Solution steps

Find the slope of

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

Compute the slope of

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

(1, - 5) : m = 3

Find the line with slope m=

3

and passing through

(1, - 5) : y = 3 x - 8

y = 3 x - 8

sin (x), 0, π

3 e^{3 x}

4 y^{^{'''}} + 12 y^{^{''}} - 16 y^{^{'}} - 48 y = 0

f (x) = 8 x^{3} - 24 x^{2} - 193 x + 48

\frac{d}{d x} (sin (x) (x^{2} - 27 x + 6))

What is the tangent of f(x)=(x-2)(x^2+4),(1,-5) ?
The tangent of f(x)=(x-2)(x^2+4),(1,-5) is y=3x-8