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

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

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

tangent of

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

y = - 12 x - 12

Solution steps

Find the slope of

f (x) = 3 x^{2} : \frac{df ( x )}{d x} = 6 x

Compute the slope of

f (x) = 3 x^{2}

(- 2, 12) : m = - 12

Find the line with slope m=

- 12

and passing through

(- 2, 12) : y = - 12 x - 12

y = - 12 x - 12

\int 8 x^{6} y^{3} + 15 x^{5} y^{4} d y

\frac{d}{d x} (x^{4} e^{2 x})

\frac{1}{2} x^{2} + 8 x + 9

y^{^{'}} = y (x y^{2} + 4)

\frac{1}{s ^{2} + 4 s + 3}

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