What is the tangent of f(x)=x^2-5x+5,\at x=4 ?

The tangent of f(x)=x^2-5x+5,\at x=4 is y=3x-11

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tangent of f(x)=x^2-5x+5,\at x=4

tangent of

f (x) = x^{2} - 5 x + 5, at x = 4

y = 3 x - 11

Solution steps

Find the tangent point:

(4, 1)

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 = 4 : m = 3

Find the line with slope m=

3

and passing through

(4, 1) : y = 3 x - 11

y = 3 x - 11

2 x^{3} + 3 x^{2} - 12 x + 4

cos (6 x)

\frac{\partial}{\partial z} (z x^{2} + y^{2})

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

x \to 0 lim (\frac{\frac{4}{x + 1} - 4}{x})

What is the tangent of f(x)=x^2-5x+5,\at x=4 ?
The tangent of f(x)=x^2-5x+5,\at x=4 is y=3x-11