What is the tangent of f(x)=8x^2+4x-6,\at x=-3 ?

The tangent of f(x)=8x^2+4x-6,\at x=-3 is y=-44x-78

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tangent of f(x)=8x^2+4x-6,\at x=-3

tangent of

f (x) = 8 x^{2} + 4 x - 6, at x = - 3

y = - 44 x - 78

Solution steps

Find the tangent point:

(- 3, 54)

Find the slope of

f (x) = 8 x^{2} + 4 x - 6 : \frac{df ( x )}{d x} = 16 x + 4

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 = - 44

Find the line with slope m=

- 44

and passing through

(- 3, 54) : y = - 44 x - 78

y = - 44 x - 78

\int \frac{u - 2 u ^{2}}{u} d u

\int a e^{x} d x

\int \frac{8}{e ^{x} - 2} d x

\int_{- \infty}^{15} r e^{\frac{r}{5}} d r

\frac{d}{d x} (3 x^{2} + 8)

What is the tangent of f(x)=8x^2+4x-6,\at x=-3 ?
The tangent of f(x)=8x^2+4x-6,\at x=-3 is y=-44x-78