What is the tangent of f(x)=9x-x^2,(1,8) ?

The tangent of f(x)=9x-x^2,(1,8) is y=7x+1

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tangent of f(x)=9x-x^2,(1,8)

tangent of

f (x) = 9 x - x^{2}, (1, 8)

y = 7 x + 1

Solution steps

Find the slope of

f (x) = 9 x - x^{2} : \frac{df ( x )}{d x} = 9 - 2 x

Compute the slope of

f (x) = 9 x - x^{2}

(1, 8) : m = 7

Find the line with slope m=

7

and passing through

(1, 8) : y = 7 x + 1

y = 7 x + 1

\frac{s + 3}{s ( s + 2 )}

x \to 2 lim (\frac{2 x - 4}{3 - 2 x})

\frac{d}{d x} (ln tan (\frac{x}{2}) + C)

\frac{d ^{2}}{d x ^{2}} (3 sec (x))

\int 3 cos (x) d x

What is the tangent of f(x)=9x-x^2,(1,8) ?
The tangent of f(x)=9x-x^2,(1,8) is y=7x+1