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

The tangent of f(x)=-4x^2-2x+3,\at x=-3 is y=22x+39

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tangent of f(x)=-4x^2-2x+3,\at x=-3

tangent of

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

y = 22 x + 39

Solution steps

Find the tangent point:

(- 3, - 27)

Find the slope of

f (x) = - 4 x^{2} - 2 x + 3 : \frac{df ( x )}{d x} = - 8 x - 2

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 = 22

Find the line with slope m=

22

and passing through

(- 3, - 27) : y = 22 x + 39

y = 22 x + 39

t \to 1 lim (t + 5)

\frac{d}{d t} (3^{t})

\frac{d y}{d x} = 1 - (10 x - y + 4)^{2}

3 y^{^{''}} + 3 y^{^{'}} - 2 y = 0

e^{(x^{2})}

What is the tangent of f(x)=-4x^2-2x+3,\at x=-3 ?
The tangent of f(x)=-4x^2-2x+3,\at x=-3 is y=22x+39