What is the tangent of f(t)=-16t^2+32t+240,\at t=240 ?

The tangent of f(t)=-16t^2+32t+240,\at t=240 is y=-7648t+921840

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tangent of f(t)=-16t^2+32t+240,\at t=240

tangent of

f (t) = - 16 t^{2} + 32 t + 240, at t = 240

y = - 7648 t + 921840

Solution steps

Find the tangent point:

(240, - 913680)

Find the slope of

f (t) = - 16 t^{2} + 32 t + 240 : \frac{df ( t )}{d t} = - 32 t + 32

Compute the slope of

f (t) = - 16 t^{2} + 32 t + 240

t = 240 : m = - 7648

Find the line with slope m=

- 7648

and passing through

(240, - 913680) : y = - 7648 t + 921840

y = - 7648 t + 921840

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\frac{\partial}{\partial x} (\frac{90}{6 + x ^{2} + y ^{2}})

\int cot (θ) d θ

\int (x (3 x - x) + (e^{e x} + 8)) d x

What is the tangent of f(t)=-16t^2+32t+240,\at t=240 ?
The tangent of f(t)=-16t^2+32t+240,\at t=240 is y=-7648t+921840