What is the tangent of f(x)=x^2-3x+2 ?

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

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tangent of f(x)=x^2-3x+2

tangent of

f (x) = 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

f (x) = x^{2} - 3 x + 2 : \frac{df ( x )}{d x} = 2 x - 3

Compute the slope of

f (x) = 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

\int_{0}^{1} 7 x e^{- x^{2}} d x

\frac{\partial}{\partial x} (x y^{2} + 7)

\frac{d}{d x} (\frac{6 x ^{4}}{x ^{2} - 3})

f (x) = cos (2 \cdot x) \cdot sin (5 \cdot x)

\frac{d}{d x} (\frac{x ^{4} - 1}{x ^{4} + 1})

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