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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 5

Find the line with slope m=

- 5

and passing through

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

y = - 5 x + 5

x \to 9 lim (x)

d er i v a t i v e \frac{6 x ^{2} + 4 x + 4}{x}

\int_{- 1}^{4} 4 - x^{2} + 3 x d x

s l o p e 6 x^{2} + 3 x y + y^{2} = 16, (1, 2)

\frac{d}{d x} (2 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