What is the tangent of f(x)=8x^2+4,\at x=-7 ?

The tangent of f(x)=8x^2+4,\at x=-7 is y=-112x-388

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tangent of f(x)=8x^2+4,\at x=-7

tangent of

f (x) = 8 x^{2} + 4, at x = - 7

y = - 112 x - 388

Solution steps

Find the tangent point:

(- 7, 396)

Find the slope of

f (x) = 8 x^{2} + 4 : \frac{df ( x )}{d x} = 16 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 = - 112

Find the line with slope m=

- 112

and passing through

(- 7, 396) : y = - 112 x - 388

y = - 112 x - 388

\frac{d}{d x} (x^{3} + 3 x^{2} + 6 x)

y = ln (\frac{cos ( x )}{cos ( x ) - 1})

\int 3 x^{9} e^{- x^{5}} d x

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

\int_{- \infty}^{0} \frac{1}{x ^{2}} d x

What is the tangent of f(x)=8x^2+4,\at x=-7 ?
The tangent of f(x)=8x^2+4,\at x=-7 is y=-112x-388