What is the tangent of f(x)=x^2-5x+5,(6,11) ?

The tangent of f(x)=x^2-5x+5,(6,11) is y=7x-31

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tangent of f(x)=x^2-5x+5,(6,11)

tangent of

f (x) = x^{2} - 5 x + 5, (6, 11)

y = 7 x - 31

Solution steps

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

(6, 11) : m = 7

Find the line with slope m=

7

and passing through

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

y = 7 x - 31

\frac{\partial}{\partial s} (s + t)

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

\int \frac{2 x}{4 - x ^{4}} d x

\frac{d}{d x} (ln (x^{3}))

\frac{d}{d x} ((\frac{1}{2} x)^{5})

What is the tangent of f(x)=x^2-5x+5,(6,11) ?
The tangent of f(x)=x^2-5x+5,(6,11) is y=7x-31