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

The tangent of g(x)=2x^2+8x-3,\at x=-3 is y=-4x-21

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tangent of g(x)=2x^2+8x-3,\at x=-3

tangent of

g (x) = 2 x^{2} + 8 x - 3, at x = - 3

y = - 4 x - 21

Solution steps

Find the tangent point:

(- 3, - 9)

Find the slope of

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

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 = - 4

Find the line with slope m=

- 4

and passing through

(- 3, - 9) : y = - 4 x - 21

y = - 4 x - 21

\int \frac{1}{3 - 2 cos ( x )} d x

\int_{1}^{16} \frac{1}{6 - x} d x

\int \frac{x ^{3} + x ^{2} - x - 3}{x ^{2} - 3} d x

\frac{cot ( t )}{e ^{t}}

f (x) = \frac{( x ^{2} )}{x + 9}, at x = 3

What is the tangent of g(x)=2x^2+8x-3,\at x=-3 ?
The tangent of g(x)=2x^2+8x-3,\at x=-3 is y=-4x-21