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

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

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tangent of y=x^2-3x+2

tangent of

y = x^{2} - 3 x + 2

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

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} + 2)

Find the slope of

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

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

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