What is the tangent of f(x)=6x-x^2,\at x=4 ?

The tangent of f(x)=6x-x^2,\at x=4 is y=-2x+16

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tangent of f(x)=6x-x^2,\at x=4

tangent of

f (x) = 6 x - x^{2}, at x = 4

y = - 2 x + 16

Solution steps

Find the tangent point:

(4, 8)

Find the slope of

f (x) = 6 x - x^{2} : \frac{df ( x )}{d x} = 6 - 2 x

Compute the slope of

f (x) = 6 x - x^{2}

x = 4 : m = - 2

Find the line with slope m=

- 2

and passing through

(4, 8) : y = - 2 x + 16

y = - 2 x + 16

\frac{d}{d x} ((x - 5) sin (x))

n = 0 \sum \infty \frac{1}{1. 5 ^{n}}

\int - x^{2} - 2 x d x

\int \frac{1}{e ^{6 x}} d x

\frac{d P}{d t} = 50 P - \frac{1}{10} P

What is the tangent of f(x)=6x-x^2,\at x=4 ?
The tangent of f(x)=6x-x^2,\at x=4 is y=-2x+16