What is the tangent of f(x)=-3x^2,\at x=1 ?

The tangent of f(x)=-3x^2,\at x=1 is y=-6x+3

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tangent of f(x)=-3x^2,\at x=1

tangent of

f (x) = - 3 x^{2}, at x = 1

y = - 6 x + 3

Solution steps

Find the tangent point:

(1, - 3)

Find the slope of

f (x) = - 3 x^{2} : \frac{df ( x )}{d x} = - 6 x

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 = - 6

Find the line with slope m=

- 6

and passing through

(1, - 3) : y = - 6 x + 3

y = - 6 x + 3

5 x^{3} - 3 x^{5}

(x y^{2}) \frac{d y}{d x} = y^{3} - x^{3}

\int x^{2} x - 2 d x

\int_{1}^{4} - 9 t ln (t) d t

x \to 0 - lim (\frac{x + 16}{x})

What is the tangent of f(x)=-3x^2,\at x=1 ?
The tangent of f(x)=-3x^2,\at x=1 is y=-6x+3