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

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

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

tangent of

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

y = - x + 1

Solution steps

Find the tangent point:

(1, 0)

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 = 1 : m = - 1

Find the line with slope m=

- 1

and passing through

(1, 0) : y = - x + 1

y = - x + 1

(x + 9) (9 x + 7)

\int x 7 - x^{2} d x

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

f (x) = cos (7 x)

\frac{x ^{2}}{5 ( 1 - x )}

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