What is the tangent of f(x)=-3x^2-5,\at x=0 ?

The tangent of f(x)=-3x^2-5,\at x=0 is y=-5

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tangent of f(x)=-3x^2-5,\at x=0

tangent of

f (x) = - 3 x^{2} - 5, at x = 0

y = - 5

Solution steps

Find the tangent point:

(0, - 5)

Find the slope of

f (x) = - 3 x^{2} - 5 : \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 = 0

Find the line with slope m=

0

and passing through

(0, - 5) : y = - 5

y = - 5

\frac{d}{d x} (\frac{x - 1}{x - 2})

\int \frac{tan ( θ )}{sec ^{2} ( θ )} d θ

\frac{8}{x}

\frac{d}{d x} (a (1 - cos^{2} (\frac{x}{2}))^{2})

(x - 2) (x - 7) (x + 1)

What is the tangent of f(x)=-3x^2-5,\at x=0 ?
The tangent of f(x)=-3x^2-5,\at x=0 is y=-5