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

The tangent of f(x)=-5x^2,\at x=2 is y=-20x+20

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

tangent of

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

y = - 20 x + 20

Solution steps

Find the tangent point:

(2, - 20)

Find the slope of

f (x) = - 5 x^{2} : \frac{df ( x )}{d x} = - 10 x

Compute the slope of

f (x) = - 5 x^{2}

x = 2 : m = - 20

Find the line with slope m=

- 20

and passing through

(2, - 20) : y = - 20 x + 20

y = - 20 x + 20

x \to 8 lim (x^{3} + 1)

\int_{3}^{5} \frac{y ^{2}}{5} + \frac{5}{4 y ^{2}} d y

\int sin (arccos (x)) d x

\int 3 e^{- t} d t

5 x y \frac{d y}{d x} = 4

What is the tangent of f(x)=-5x^2,\at x=2 ?
The tangent of f(x)=-5x^2,\at x=2 is y=-20x+20