What is the tangent of f(x)=x^2-5x+5,\at x=0 ?

The tangent of f(x)=x^2-5x+5,\at x=0 is y=-5x+5

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tangent of f(x)=x^2-5x+5,\at x=0

tangent of

f (x) = x^{2} - 5 x + 5, at x = 0

y = - 5 x + 5

Solution steps

Find the tangent point:

(0, 5)

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 = 0 : m = - 5

Find the line with slope m=

- 5

and passing through

(0, 5) : y = - 5 x + 5

y = - 5 x + 5

\int x^{2} + 4 x d x

y^{^{''}} + y = - sin (2 x)

\frac{d}{d y} (arctan (\frac{x}{y}))

\frac{d}{d x} (\frac{x}{( 1 + 4 x ) ^{2}})

\int \frac{12}{4 x ^{2} + 5 x + 1} d x

What is the tangent of f(x)=x^2-5x+5,\at x=0 ?
The tangent of f(x)=x^2-5x+5,\at x=0 is y=-5x+5