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

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

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

tangent of

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

y = 4 x + 1

Solution steps

Find the slope of

y = 6 x - x^{2} : \frac{d y}{d x} = 6 - 2 x

Compute the slope of

y = 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

\int \frac{x + 2}{x ^{2} + 9} d x

\int \frac{6}{x - 5} d x

\frac{\partial}{\partial y} (3 x^{2} y)

\int_{0}^{12} 1 + \frac{x}{4} d x

y = 14 (sin (x) + cos (x))