What is the tangent of f(x)=5x-4x^2,(-1,-9) ?

The tangent of f(x)=5x-4x^2,(-1,-9) is y=13x+4

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tangent of f(x)=5x-4x^2,(-1,-9)

tangent of

f (x) = 5 x - 4 x^{2}, (- 1, - 9)

y = 13 x + 4

Solution steps

Find the slope of

f (x) = 5 x - 4 x^{2} : \frac{df ( x )}{d x} = 5 - 8 x

Compute the slope of

f (x) = 5 x - 4 x^{2}

(- 1, - 9) : m = 13

Find the line with slope m=

13

and passing through

(- 1, - 9) : y = 13 x + 4

y = 13 x + 4

n = 0 \sum \infty \frac{5}{2 + 3 ^{n}}

\int \frac{csc ( x ) cot ( x )}{2 - csc ( x )} d x

\int e^{- x} (9 + e^{- x}) d x

y \to 6 lim (\frac{2 y ^{2} - 15 y + 18}{3 y ^{2} - 17 y - 6})

\int x^{\frac{1}{7}} d x

What is the tangent of f(x)=5x-4x^2,(-1,-9) ?
The tangent of f(x)=5x-4x^2,(-1,-9) is y=13x+4