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

The tangent of f(x)=2x^2-4,\at x=8 is y=32x-132

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

tangent of

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

y = 32 x - 132

Solution steps

Find the tangent point:

(8, 124)

Find the slope of

f (x) = 2 x^{2} - 4 : \frac{df ( x )}{d x} = 4 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 = 32

Find the line with slope m=

32

and passing through

(8, 124) : y = 32 x - 132

y = 32 x - 132

\frac{\partial}{\partial x} (z arctan (\frac{x}{y}))

\int 6 (x + 2) \cdot ln (x + 2) d x

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

y = 5^{t a n (5 x)}

\int \frac{4}{y} d y

What is the tangent of f(x)=2x^2-4,\at x=8 ?
The tangent of f(x)=2x^2-4,\at x=8 is y=32x-132