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

The tangent of f(x)=-3x^2+12,\at x=3 is y=-18x+39

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

tangent of

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

y = - 18 x + 39

Solution steps

Find the tangent point:

(3, - 15)

Find the slope of

f (x) = - 3 x^{2} + 12 : \frac{df ( x )}{d x} = - 6 x

Compute the slope of

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

x = 3 : m = - 18

Find the line with slope m=

- 18

and passing through

(3, - 15) : y = - 18 x + 39

y = - 18 x + 39

8 x^{2} y + y^{^{'}} = 10 x^{2}

4 x^{3} - 33 x^{2} + 84 x - 60

\frac{d}{d x} (2 x 16 - x^{2})

\frac{\partial}{\partial x} (2 x^{2} + y^{3})

f (x) = \frac{2 x ^{\frac{3}{2}} - x ^{2} + 4}{x}

What is the tangent of f(x)=-3x^2+12,\at x=3 ?
The tangent of f(x)=-3x^2+12,\at x=3 is y=-18x+39