What is the tangent of y=x^3-3x+2,(4,54) ?

The tangent of y=x^3-3x+2,(4,54) is y=45x-126

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tangent of y=x^3-3x+2,(4,54)

tangent of

y = x^{3} - 3 x + 2, (4, 54)

y = 45 x - 126

Solution steps

Find the slope of

y = x^{3} - 3 x + 2 : \frac{d y}{d x} = 3 x^{2} - 3

Compute the slope of

y = x^{3} - 3 x + 2

(4, 54) : m = 45

Find the line with slope m=

45

and passing through

(4, 54) : y = 45 x - 126

y = 45 x - 126

\frac{s ^{2} + 12}{s ^{4} - 16}

\frac{d}{d x} (\frac{x ^{2}}{2 - x})

x \to 0 lim (sin (x) ln (x))

\int \frac{29 x ^{2}}{x ^{2} - 4} d x

y = x^{2} + 2 x, y = x + 2

What is the tangent of y=x^3-3x+2,(4,54) ?
The tangent of y=x^3-3x+2,(4,54) is y=45x-126