What is the tangent of f(x)=6x-x^2,\at x=-1 ?

The tangent of f(x)=6x-x^2,\at x=-1 is y=8x+1

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tangent of f(x)=6x-x^2,\at x=-1

tangent of

f (x) = 6 x - x^{2}, at x = - 1

y = 8 x + 1

Solution steps

Find the tangent point:

(- 1, - 7)

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 = - 1 : m = 8

Find the line with slope m=

8

and passing through

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

y = 8 x + 1

f^{^{'}} (θ) = cos (θ) - sin (θ)

\frac{\partial}{\partial y} (4 x^{3} y^{3} + 4)

\int_{- 1}^{3} 4 - x^{2} d x

y = \frac{3}{x}, y = 7 x, y = \frac{1}{7} x

f (x) = \frac{( x ^{2} - 6 x )}{x + 1}

What is the tangent of f(x)=6x-x^2,\at x=-1 ?
The tangent of f(x)=6x-x^2,\at x=-1 is y=8x+1