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

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

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

tangent of

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

y = x - 2

Solution steps

Find the tangent point:

(2, 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 = 2 : m = 1

Find the line with slope m=

1

and passing through

(2, 0) : y = x - 2

y = x - 2

\int \frac{2 x - 3}{6} d x

ln (x^{2})

\frac{1}{( s + 3 )}

\int \frac{1}{( 2 x - 3 ) \cdot ( 3 x ^{2} + x )} d x

\frac{d}{d x} (\frac{x ^{2}}{x ^{4}})

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