What is the tangent of f(x)=110x-16x^2,\at 3 ?

The tangent of f(x)=110x-16x^2,\at 3 is y=(110-32a_{0})x+16a_{0}^2

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tangent of f(x)=110x-16x^2,\at 3

tangent of

f (x) = 110 x - 16 x^{2}, at 3

y = (110 - 32 a_{0}) x + 16 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 110 a_{0} - 16 a_{0}^{2})

Find the slope of

f (x) = 110 x - 16 x^{2} : \frac{df ( x )}{d x} = 110 - 32 x

Compute the slope of

f (x) = 110 x - 16 x^{2}

x = a_{0} : m = 110 - 32 a_{0}

Find the line with slope m=

110 - 32 a_{0}

and passing through

(a_{0}, 110 a_{0} - 16 a_{0}^{2}) : y = (110 - 32 a_{0}) x + 16 a_{0}^{2}

y = (110 - 32 a_{0}) x + 16 a_{0}^{2}

7 x

\frac{d x}{d t} = (\frac{( e ^{t} x + 1 )}{1 - e ^{t}})

\int \frac{x - 3}{( x + 4 ) ( x + 5 )} d x

csc (θ)

\int_{π}^{2 π} x sin (\frac{x}{2}) d x

What is the tangent of f(x)=110x-16x^2,\at 3 ?
The tangent of f(x)=110x-16x^2,\at 3 is y=(110-32a_{0})x+16a_{0}^2