What is the tangent of f(x)=6x-x^2,(1,5) ?

The tangent of f(x)=6x-x^2,(1,5) is y=4x+1

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tangent of f(x)=6x-x^2,(1,5)

tangent of

f (x) = 6 x - x^{2}, (1, 5)

y = 4 x + 1

Solution steps

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}

(1, 5) : m = 4

Find the line with slope m=

4

and passing through

(1, 5) : y = 4 x + 1

y = 4 x + 1

f (x) = \frac{1}{x + 8}

y^{^{'}} - y = x e^{2 x} + 1

\frac{d}{d x} (5^{2 x^{3}})

x y^{^{'}} = 4 x^{2} - 2 y, y (1) = 3

y^{2} - 27 x = 0, x + y = 0

What is the tangent of f(x)=6x-x^2,(1,5) ?
The tangent of f(x)=6x-x^2,(1,5) is y=4x+1