What is the tangent of y=x^2-3x+1 ?

The tangent of y=x^2-3x+1 is y=(2a_{0}-3)x-a_{0}^2+1

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tangent of y=x^2-3x+1

tangent of

y = x^{2} - 3 x + 1

y = (2 a_{0} - 3) x - a_{0}^{2} + 1

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} + 1)

Find the slope of

y = x^{2} - 3 x + 1 : \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} + 1) : y = (2 a_{0} - 3) x - a_{0}^{2} + 1

y = (2 a_{0} - 3) x - a_{0}^{2} + 1

\frac{\partial}{\partial x} (4 (x - y) e^{4 y + 4 x^{2}})

x^{^{''}} + 3 x^{^{'}} - 10 x = 6 e^{4 t}

\frac{\partial}{\partial x} (2 y x^{2} - 4)

\int_{2}^{\infty} \frac{1}{3 x} d x

r^{2} - 13 x^{2}

What is the tangent of y=x^2-3x+1 ?
The tangent of y=x^2-3x+1 is y=(2a_{0}-3)x-a_{0}^2+1