What is the tangent of x^2-x/5 ?

The tangent of x^2-x/5 is y=(2a_{0}-1/5)x-a_{0}^2

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tangent of x^2-x/5

tangent of

x^{2} - \frac{x}{5}

y = (2 a_{0} - \frac{1}{5}) x - a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - \frac{a _{0}}{5})

Find the slope of

y = x^{2} - \frac{x}{5} : \frac{d y}{d x} = 2 x - \frac{1}{5}

Compute the slope of

y = x^{2} - \frac{x}{5}

x = a_{0} : m = 2 a_{0} - \frac{1}{5}

Find the line with slope m=

2 a_{0} - \frac{1}{5}

and passing through

(a_{0}, a_{0}^{2} - \frac{a _{0}}{5}) : y = (2 a_{0} - \frac{1}{5}) x - a_{0}^{2}

y = (2 a_{0} - \frac{1}{5}) x - a_{0}^{2}

\frac{d y}{d x} = y (5 - y), y = 6

\frac{d r}{d t} = \frac{0.15 r}{0.5} + 0.1

\int sin^{2} (x) cos^{7} (x) d x

\int_{1}^{\infty} 7 e^{- 7 x} d x

y^{2} = x^{3} + 3 x^{2}, (1, 2)

What is the tangent of x^2-x/5 ?
The tangent of x^2-x/5 is y=(2a_{0}-1/5)x-a_{0}^2