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

The tangent of y=t^3-4t-3,(2,-3) is y=8t-19

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tangent of y=t^3-4t-3,(2,-3)

tangent of

y = t^{3} - 4 t - 3, (2, - 3)

y = 8 t - 19

Solution steps

Find the slope of

y = t^{3} - 4 t - 3 : \frac{d y}{d t} = 3 t^{2} - 4

Compute the slope of

y = t^{3} - 4 t - 3

(2, - 3) : m = 8

Find the line with slope m=

8

and passing through

(2, - 3) : y = 8 t - 19

y = 8 t - 19

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

\int (y e^{y}) d y

\int_{0}^{5} π (2 5 y)^{2} d y

\int \frac{e ^{x}}{36 - e ^{2 x}} d x

\int x 2 - x d x

What is the tangent of y=t^3-4t-3,(2,-3) ?
The tangent of y=t^3-4t-3,(2,-3) is y=8t-19