What is the tangent of f(x)=110x-16x^2,\at 3.438 ?

The tangent of f(x)=110x-16x^2,\at 3.438 is y=(110-32a_{0})x+16a_{0}^2

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tangent of f(x)=110x-16x^2,\at 3.438

tangent of

f (x) = 110 x - 16 x^{2}, at 3.438

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}

\frac{x ^{4} - 16 x ^{2} - 16}{( x ^{2} - 4 ) ^{2}}

x y d y - (x^{2} + y^{2}) d x = 0

n = 1 \sum \infty \frac{1}{( 0.8 ) ^{n}}

\int_{0}^{2} x^{2} 4 - x^{2} d x

f (x) = \frac{- 8 x}{x ^{2} + 1}

What is the tangent of f(x)=110x-16x^2,\at 3.438 ?
The tangent of f(x)=110x-16x^2,\at 3.438 is y=(110-32a_{0})x+16a_{0}^2