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

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

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tangent of x^2-3x+2

tangent of

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

Compute the slope of

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

x = a_{0} : 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

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

\int \frac{1}{sin ( x ) - cos ( x ) + 2} d x

\int 8 x sin (x) d x

\frac{d}{d x} y = (x + y + 3)^{2}

\int 5 e^{6 - 3 x} d x

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