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-6

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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 - 6

Solution steps

Find the tangent point:

(- 3, 12)

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

x = - 3 : m = - 6

Find the line with slope m=

- 6

and passing through

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

y = - 6 x - 6

(e^{x} + 1)^{2}

x \to 0 lim (\frac{x ^{2} + 3}{x ^{4}})

\int_{0}^{1} 5 cos (x^{2}) d x

x \to 0 lim (\frac{( 6 ^{x} - 8 ^{x} )}{x})

\int cos (t) + sin (t) d t

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-6