What is the tangent of y=(t-5)(t^2-3),(2,-3) ?

The tangent of y=(t-5)(t^2-3),(2,-3) is y=-11x+19

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tangent of y=(t-5)(t^2-3),(2,-3)

tangent of

y = (x - 5) (x^{2} - 3), (2, - 3)

y = - 11 x + 19

Solution steps

Find the slope of

y = (x - 5) (x^{2} - 3) : \frac{d y}{d x} = 3 x^{2} - 10 x - 3

Compute the slope of

y = (x - 5) (x^{2} - 3)

(2, - 3) : m = - 11

Find the line with slope m=

- 11

and passing through

(2, - 3) : y = - 11 x + 19

y = - 11 x + 19

x \to 0 lim (x^{18} ln (x))

\frac{d}{d x} ((x^{3} - 1)^{100})

23 t

x^{^{''}} - 3 x^{^{'}} + 2 x = 0

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

What is the tangent of y=(t-5)(t^2-3),(2,-3) ?
The tangent of y=(t-5)(t^2-3),(2,-3) is y=-11x+19