What is the tangent of y=(s-4)(s^2-3),(1,6) ?

The tangent of y=(s-4)(s^2-3),(1,6) is y=-8x+14

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tangent of y=(s-4)(s^2-3),(1,6)

tangent of

y = (x - 4) (x^{2} - 3), (1, 6)

y = - 8 x + 14

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 6) : m = - 8

Find the line with slope m=

- 8

and passing through

(1, 6) : y = - 8 x + 14

y = - 8 x + 14

t sin (t)

\frac{d}{d x} (\frac{x ^{4}}{4})

\int_{0}^{\frac{π}{20}} cos (5 x) d x

x \to 0 + lim (\frac{7}{x ^{3} + 2 x ^{2}})

\frac{d}{d x} (e^{2 x + l n (x)})

What is the tangent of y=(s-4)(s^2-3),(1,6) ?
The tangent of y=(s-4)(s^2-3),(1,6) is y=-8x+14