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

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

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

tangent of

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

y = 13 x + 4

Solution steps

Find the tangent point:

(- 1, - 9)

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}

x = - 1 : m = 13

Find the line with slope m=

13

and passing through

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

y = 13 x + 4

\int_{0}^{3} (2^{x}) d x

y^{^{''}} + 2 y^{^{'}} + 4 y = 50 cos (x)

\int 2 e^{2 x} sin (e^{2 x}) d x

x y^{^{'}} + y = - 6 x y^{2}

\frac{3}{2} x^{\frac{2}{3}}

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