What is the tangent of 128t-16t^2,\at [4,5] ?

The tangent of 128t-16t^2,\at [4,5] is y=(128-32a_{0})t+16a_{0}^2

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tangent of 128t-16t^2,\at [4,5]

tangent of

128 t - 16 t^{2}, at [4, 5]

y = (128 - 32 a_{0}) t + 16 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

t = a_{0}

Find the tangent point:

(a_{0}, 128 a_{0} - 16 a_{0}^{2})

Find the slope of

y = 128 t - 16 t^{2} : \frac{d y}{d t} = 128 - 32 t

Compute the slope of

y = 128 t - 16 t^{2}

t = a_{0} : m = 128 - 32 a_{0}

Find the line with slope m=

128 - 32 a_{0}

and passing through

(a_{0}, 128 a_{0} - 16 a_{0}^{2}) : y = (128 - 32 a_{0}) t + 16 a_{0}^{2}

y = (128 - 32 a_{0}) t + 16 a_{0}^{2}

x \to 1 - lim ((1 - x)^{l n (x)})

\frac{1 + x}{1 - x}

\frac{d y}{d x} + 5 y = 9

\int arccot (14 y) d y

(2 x + 1) y^{^{'}} = 4 x + 2 y

What is the tangent of 128t-16t^2,\at [4,5] ?
The tangent of 128t-16t^2,\at [4,5] is y=(128-32a_{0})t+16a_{0}^2