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

The tangent of f(x)=110x-16x^2,\at x=3.438 is y=-0.016x+189.117504

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

tangent of

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

y = - 0.016 x + 189.117504

Solution steps

Find the tangent point:

(3.438, 189.062496)

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 = 3.438 : m = - 0.016

Find the line with slope m=

- 0.016

and passing through

(3.438, 189.062496) : y = - 0.016 x + 189.117504

y = - 0.016 x + 189.117504

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x \to - \infty lim (\frac{2 x ^{2} + 3}{2 x + 3})

x \to - 1 lim (3)

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What is the tangent of f(x)=110x-16x^2,\at x=3.438 ?
The tangent of f(x)=110x-16x^2,\at x=3.438 is y=-0.016x+189.117504