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

The tangent of f(x)=2x^2+8x,\at x=4 is y=24x-32

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

tangent of

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

y = 24 x - 32

Solution steps

Find the tangent point:

(4, 64)

Find the slope of

f (x) = 2 x^{2} + 8 x : \frac{df ( 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 = 24

Find the line with slope m=

24

and passing through

(4, 64) : y = 24 x - 32

y = 24 x - 32

n \to 0 lim (sin (\frac{1}{n}))

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

\frac{d ^{2}}{d x ^{2}} (x^{2} + y^{2})

f (x) = 7 x + 4

f (x) = - 2 sin (x), at x = - π

What is the tangent of f(x)=2x^2+8x,\at x=4 ?
The tangent of f(x)=2x^2+8x,\at x=4 is y=24x-32