What is the tangent of f(x)=8x^2+3x,\at x=-45 ?

The tangent of f(x)=8x^2+3x,\at x=-45 is y=-717x-16200

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tangent of f(x)=8x^2+3x,\at x=-45

tangent of

f (x) = 8 x^{2} + 3 x, at x = - 45

y = - 717 x - 16200

Solution steps

Find the tangent point:

(- 45, 16065)

Find the slope of

f (x) = 8 x^{2} + 3 x : \frac{df ( x )}{d x} = 16 x + 3

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 = - 717

Find the line with slope m=

- 717

and passing through

(- 45, 16065) : y = - 717 x - 16200

y = - 717 x - 16200

\frac{d y}{d x} = 2 sec^{2} (x) e^{2 y}, y (0) = 0

x \to 3 lim (\frac{x - 2 - 1}{x - 3})

\int \frac{1}{x + 4 + x + 3} d x

y^{^{''}} + 4 y^{^{'}} + 8 y = 4 cos (2 t) + 2 sin (2 t)

\frac{d}{d x} (\frac{x ^{8}}{8 ( 1 - x ^{2} ) ^{4}})

What is the tangent of f(x)=8x^2+3x,\at x=-45 ?
The tangent of f(x)=8x^2+3x,\at x=-45 is y=-717x-16200