What is the tangent of f(x)=x^2-5x+5,\at x=6 ?

The tangent of f(x)=x^2-5x+5,\at x=6 is y=7x-31

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tangent of f(x)=x^2-5x+5,\at x=6

tangent of

f (x) = x^{2} - 5 x + 5, at x = 6

y = 7 x - 31

Solution steps

Find the tangent point:

(6, 11)

Find the slope of

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

Compute the slope of

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

x = 6 : m = 7

Find the line with slope m=

7

and passing through

(6, 11) : y = 7 x - 31

y = 7 x - 31

x^{2} cos (x), 0.1

f (x) = - 25 x^{2} + 150 x - 182

\int 4 e^{- 0.2 x} d x

\int csc^{3} (2 x) cot^{3} (2 x) d x

19 x e^{- x}

What is the tangent of f(x)=x^2-5x+5,\at x=6 ?
The tangent of f(x)=x^2-5x+5,\at x=6 is y=7x-31