What is the tangent of f(x)=4-3x^2,(-2,-8) ?

The tangent of f(x)=4-3x^2,(-2,-8) is y=12x+16

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tangent of f(x)=4-3x^2,(-2,-8)

tangent of

f (x) = 4 - 3 x^{2}, (- 2, - 8)

y = 12 x + 16

Solution steps

Find the slope of

f (x) = 4 - 3 x^{2} : \frac{df ( x )}{d x} = - 6 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 = 12

Find the line with slope m=

12

and passing through

(- 2, - 8) : y = 12 x + 16

y = 12 x + 16

What is the tangent of f(x)=4-3x^2,(-2,-8) ?
The tangent of f(x)=4-3x^2,(-2,-8) is y=12x+16