What is the tangent of x^2-9x ?

The tangent of x^2-9x is y=(2a_{0}-9)x-a_{0}^2

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tangent of x^2-9x

tangent of

x^{2} - 9 x

y = (2 a_{0} - 9) 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} - 9 a_{0})

Find the slope of

y = x^{2} - 9 x : \frac{d y}{d x} = 2 x - 9

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2 a_{0} - 9

Find the line with slope m=

2 a_{0} - 9

and passing through

(a_{0}, a_{0}^{2} - 9 a_{0}) : y = (2 a_{0} - 9) x - a_{0}^{2}

y = (2 a_{0} - 9) x - a_{0}^{2}

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