What is the tangent of f(x)=3x^2+5x-4,\at x=-0.6667 ?

The tangent of f(x)=3x^2+5x-4,\at x=-0.6667 is y=0.9998x-5.33346667

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tangent of f(x)=3x^2+5x-4,\at x=-0.6667

tangent of

f (x) = 3 x^{2} + 5 x - 4, at x = - 0.6667

y = 0.9998 x - 5.33346667

Solution steps

Find the tangent point:

(- 0.6667, - 6.00003333)

Find the slope of

f (x) = 3 x^{2} + 5 x - 4 : \frac{df ( x )}{d x} = 6 x + 5

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 = 0.9998

Find the line with slope m=

0.9998

and passing through

(- 0.6667, - 6.00003333) : y = 0.9998 x - 5.33346667

y = 0.9998 x - 5.33346667

y = u + \frac{2}{u}, u = (2 x + 1)^{3}, x = 0

y^{^{''}} + y = (t^{2} + t) \cdot sin (t)

f (x) = 0.5 x^{2} + 2 x - 3, at x = - 3

\int \frac{1}{z ^{3} z ^{2} - 4} d z

y^{^{'}} + 3 y = 2 e^{- 3 t}, y (0) = 2

What is the tangent of f(x)=3x^2+5x-4,\at x=-0.6667 ?
The tangent of f(x)=3x^2+5x-4,\at x=-0.6667 is y=0.9998x-5.33346667