What is the tangent of f(x)=x^2-10x+2,\at x=0 ?

The tangent of f(x)=x^2-10x+2,\at x=0 is y=-10x+2

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tangent of f(x)=x^2-10x+2,\at x=0

tangent of

f (x) = x^{2} - 10 x + 2, at x = 0

y = - 10 x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

f (x) = x^{2} - 10 x + 2 : \frac{df ( x )}{d x} = 2 x - 10

Compute the slope of

f (x) = x^{2} - 10 x + 2

x = 0 : m = - 10

Find the line with slope m=

- 10

and passing through

(0, 2) : y = - 10 x + 2

y = - 10 x + 2

x \to 0 lim (csc (x) - cot (x))

\int 0.1 e^{- 0.1 x} d x

\frac{d y}{d t} = sin (3 t) - 2 y

x \to 9 + lim (\frac{5}{x - 9})

\int (5 + 3 x + 1) d x

What is the tangent of f(x)=x^2-10x+2,\at x=0 ?
The tangent of f(x)=x^2-10x+2,\at x=0 is y=-10x+2