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

The tangent of f(x)=x^2-3x+1,\at x=3 is y=3x-8

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

tangent of

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

y = 3 x - 8

Solution steps

Find the tangent point:

(3, 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 = 3 : m = 3

Find the line with slope m=

3

and passing through

(3, 1) : y = 3 x - 8

y = 3 x - 8

n = 2 \sum \infty \frac{1}{( 3 n - 2 ) ^{5}}

f (x) = - x^{3} + 2 x^{2} - 4, at x = - 1

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

(3 x^{7} - 4) (2 x - 5 x^{2})

What is the tangent of f(x)=x^2-3x+1,\at x=3 ?
The tangent of f(x)=x^2-3x+1,\at x=3 is y=3x-8