What is the tangent of x^2-6x ?

The tangent of x^2-6x is y=(2a_{0}-6)x-a_{0}^2

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tangent of x^2-6x

tangent of

x^{2} - 6 x

y = (2 a_{0} - 6) 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} - 6 a_{0})

Find the slope of

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

Compute the slope of

y = x^{2} - 6 x

x = a_{0} : m = 2 a_{0} - 6

Find the line with slope m=

2 a_{0} - 6

and passing through

(a_{0}, a_{0}^{2} - 6 a_{0}) : y = (2 a_{0} - 6) x - a_{0}^{2}

y = (2 a_{0} - 6) x - a_{0}^{2}

f (x) = \frac{3}{x - 2}

- 4 cos (3 t)

\int \frac{10800 q}{9 q ^{2} + 675} d q

\int \frac{( arctan ( x ) )}{1 + x ^{2}} d x

f (x) = x^{3}, (0, 0)