What is the tangent of f(x)=x^2-3x+5,\at x=2 ?

The tangent of f(x)=x^2-3x+5,\at x=2 is y=x+1

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tangent of f(x)=x^2-3x+5,\at x=2

tangent of

f (x) = x^{2} - 3 x + 5, at x = 2

y = x + 1

Solution steps

Find the tangent point:

(2, 3)

Find the slope of

f (x) = x^{2} - 3 x + 5 : \frac{df ( x )}{d x} = 2 x - 3

Compute the slope of

f (x) = x^{2} - 3 x + 5

x = 2 : m = 1

Find the line with slope m=

1

and passing through

(2, 3) : y = x + 1

y = x + 1

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

\int \frac{x ^{2} - 12}{x ^{3} + 4 x} d x

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

\frac{d}{d x} (\frac{( x ^{2} - 2 e ^{x} )}{x + 3 e ^{x}})

\int \frac{1}{x 3 x} d x

What is the tangent of f(x)=x^2-3x+5,\at x=2 ?
The tangent of f(x)=x^2-3x+5,\at x=2 is y=x+1