What is the tangent of f(t)=-16t^2+32t+240,\at t=0 ?

The tangent of f(t)=-16t^2+32t+240,\at t=0 is y=32t+240

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tangent of f(t)=-16t^2+32t+240,\at t=0

tangent of

f (t) = - 16 t^{2} + 32 t + 240, at t = 0

y = 32 t + 240

Solution steps

Find the tangent point:

(0, 240)

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 = 0 : m = 32

Find the line with slope m=

32

and passing through

(0, 240) : y = 32 t + 240

y = 32 t + 240

x \to + 2 lim (\frac{x ^{2} - 4}{x - 2})

x^{2}, 4 x

y = 1 + 2 x, y = \frac{3 + x}{3}

x, \frac{x}{2}

y = \frac{1}{x}, y = \frac{1}{x ^{2}}, x = 4

What is the tangent of f(t)=-16t^2+32t+240,\at t=0 ?
The tangent of f(t)=-16t^2+32t+240,\at t=0 is y=32t+240