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

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

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tangent of f(x)=x^2+3,(-3,12)

tangent of

f (x) = x^{2} + 3, (- 3, 12)

y = - 6 x - 6

Solution steps

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, 12) : y = - 6 x - 6

y = - 6 x - 6

\frac{\partial}{\partial x} (f (x, y) (x + y))

y^{^{''}} + y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{d}{d x} (\frac{1 - 3 x - x ^{2}}{x ^{2} - 4})

\frac{\partial}{\partial x} (\frac{x}{x ^{5} - y ^{7}})

e^{- 2 a x}

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