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

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

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tangent of x^2-5x

tangent of

x^{2} - 5 x

y = (2 a_{0} - 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} - 5 a_{0})

Find the slope of

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

Compute the slope of

y = x^{2} - 5 x

x = a_{0} : m = 2 a_{0} - 5

Find the line with slope m=

2 a_{0} - 5

and passing through

(a_{0}, a_{0}^{2} - 5 a_{0}) : y = (2 a_{0} - 5) x - a_{0}^{2}

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

x y^{^{'}} + 4 y = 9 x^{5} + 2 x^{3}

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 t}

\int \frac{1}{5 + 2 t} d t

\frac{\partial}{\partial a} (\frac{a ^{3} b ^{4}}{c ^{2} d ^{4}})

2 (x - y) d x + d y = 0