What is the tangent of f(x)=x^2-3x+1,\at x=1 ?

The tangent of f(x)=x^2-3x+1,\at x=1 is y=-x

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tangent of f(x)=x^2-3x+1,\at x=1

tangent of

f (x) = x^{2} - 3 x + 1, at x = 1

y = - x

Solution steps

Find the tangent point:

(1, - 1)

Find the slope of

f (x) = x^{2} - 3 x + 1 : \frac{df ( x )}{d x} = 2 x - 3

Compute the slope of

f (x) = x^{2} - 3 x + 1

x = 1 : m = - 1

Find the line with slope m=

- 1

and passing through

(1, - 1) : y = - x

y = - x

\frac{\partial}{\partial x} (x^{2} - 4 x y + y^{3} + 4 y)

\frac{- 200 ln ( \frac{f}{40} )}{( f + 20 )}

\frac{d}{d x} (\frac{y}{1 + x ^{2}})

\int \frac{x ^{3}}{5} d x

\int_{1}^{\infty} \frac{1}{n ^{2}}

What is the tangent of f(x)=x^2-3x+1,\at x=1 ?
The tangent of f(x)=x^2-3x+1,\at x=1 is y=-x