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

The tangent of f(x)=-2x^2-1,\at x=1 is y=-4x+1

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

tangent of

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

y = - 4 x + 1

Solution steps

Find the tangent point:

(1, - 3)

Find the slope of

f (x) = - 2 x^{2} - 1 : \frac{df ( x )}{d x} = - 4 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 = - 4

Find the line with slope m=

- 4

and passing through

(1, - 3) : y = - 4 x + 1

y = - 4 x + 1

\frac{d x}{d y} = \frac{x + y}{y}

\int - 3 x^{4} + x^{3} + 5 - 5 e^{x} d x

x \to \frac{9 π}{2} - lim (tan (x))

\int sin^{3} (a) x d x

\int (\frac{1}{2 x + 1}) d x

What is the tangent of f(x)=-2x^2-1,\at x=1 ?
The tangent of f(x)=-2x^2-1,\at x=1 is y=-4x+1