What is the tangent of f(x)=x^2-3,(3,6) ?

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

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tangent of f(x)=x^2-3,(3,6)

tangent of

f (x) = x^{2} - 3, (3, 6)

y = 6 x - 12

Solution steps

Find the slope of

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

Compute the slope of

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

(3, 6) : m = 6

Find the line with slope m=

6

and passing through

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

y = 6 x - 12

\frac{d y}{d x} = 3 x e^{y}, y (0) = 0

\int_{1}^{7} \frac{2 ln ( x )}{x} d x

\int_{0}^{1} 6^{4 x} d x

y^{^{''}} - 3 y^{^{'}} = 18 t

\frac{d}{d t} (A e^{- 2 t} cos (t) + B e^{- 2 t} sin (t))

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