What is the tangent of x^2-3x+3 ?

The tangent of x^2-3x+3 is y=(2a_{0}-3)x-a_{0}^2+3

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tangent of x^2-3x+3

tangent of

x^{2} - 3 x + 3

y = (2 a_{0} - 3) x - a_{0}^{2} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 3 a_{0} + 3)

Find the slope of

y = x^{2} - 3 x + 3 : \frac{d y}{d x} = 2 x - 3

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} - 3

Find the line with slope m=

2 a_{0} - 3

and passing through

(a_{0}, a_{0}^{2} - 3 a_{0} + 3) : y = (2 a_{0} - 3) x - a_{0}^{2} + 3

y = (2 a_{0} - 3) x - a_{0}^{2} + 3

\int 2 y e^{2 z} d z

h (g^{^{'}} (x))

y = \frac{4}{x}, y = 16 x, y = \frac{1}{16} x

y = 3 x ln (x)

\frac{\partial}{\partial x} (\frac{2 x - 2 y}{2 x + 2 y})

What is the tangent of x^2-3x+3 ?
The tangent of x^2-3x+3 is y=(2a_{0}-3)x-a_{0}^2+3