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

The tangent of f(x)=x^2-3,\at x=3 is y=6x-12

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

tangent of

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

y = 6 x - 12

Solution steps

Find the tangent point:

(3, 6)

Find the slope of

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

(3, 6) : y = 6 x - 12

y = 6 x - 12

\int_{a}^{2 a} (a x - \frac{a ^{3}}{x}) d x

x \to \infty lim (4 e^{- 4 x})

\int x^{\frac{6}{5}} d x

\int_{1}^{3} 3 (2^{x + 1}) d x

\frac{d}{d x} (3 x^{4} - 2 x^{2} + 8)

What is the tangent of f(x)=x^2-3,\at x=3 ?
The tangent of f(x)=x^2-3,\at x=3 is y=6x-12