What is the tangent of f(x)=110x-16x^2,\at x=4 ?

The tangent of f(x)=110x-16x^2,\at x=4 is y=-18x+256

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tangent of f(x)=110x-16x^2,\at x=4

tangent of

f (x) = 110 x - 16 x^{2}, at x = 4

y = - 18 x + 256

Solution steps

Find the tangent point:

(4, 184)

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 = 4 : m = - 18

Find the line with slope m=

- 18

and passing through

(4, 184) : y = - 18 x + 256

y = - 18 x + 256

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

\frac{1}{x + 8}

\frac{d}{d x} (\frac{e ^{x}}{x ^{2} + 7})

\int_{1}^{\infty} \frac{5}{2 x} d x

What is the tangent of f(x)=110x-16x^2,\at x=4 ?
The tangent of f(x)=110x-16x^2,\at x=4 is y=-18x+256