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

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

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

tangent of

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

y = 2 x - 1

Solution steps

Find the tangent point:

(2, 3)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 2

Find the line with slope m=

2

and passing through

(2, 3) : y = 2 x - 1

y = 2 x - 1

x \to 0 lim (Q (x))

\int \frac{7}{3 p - 4} d p

\int e^{2 x} cos (e^{2 x}) d x

\int \frac{16 x ^{2}}{( x - 15 ) ( x + 5 ) ^{2}} d x

\int \frac{x ^{3}}{3 + x ^{2}} d x

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