What is the tangent of h(t)=2t^3,(-2,-16) ?

The tangent of h(t)=2t^3,(-2,-16) is y=24t+32

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tangent of h(t)=2t^3,(-2,-16)

tangent of

h (t) = 2 t^{3}, (- 2, - 16)

y = 24 t + 32

Solution steps

Find the slope of

h (t) = 2 t^{3} : \frac{d h ( t )}{d t} = 6 t^{2}

Compute the slope of

h (t) = 2 t^{3}

(- 2, - 16) : m = 24

Find the line with slope m=

24

and passing through

(- 2, - 16) : y = 24 t + 32

y = 24 t + 32

\int (- \frac{3000}{x ^{2}}) d x

\frac{d}{d x} (\frac{8}{x ^{2} + 6 x})

x \to 0 lim (3 e^{3 x})

\int_{0}^{1} f (x) d x

\frac{d}{d x} (\frac{( 2 + x )}{( 2 - x ) ^{5}})

What is the tangent of h(t)=2t^3,(-2,-16) ?
The tangent of h(t)=2t^3,(-2,-16) is y=24t+32