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

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 = - 8

Find the line with slope m=

- 8

and passing through

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

y = - 8 x + 14

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