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

The tangent of f(x)=x^2-8,\at x=3 is y=6x-17

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

tangent of

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

y = 6 x - 17

Solution steps

Find the tangent point:

(3, 1)

Find the slope of

f (x) = x^{2} - 8 : \frac{df ( x )}{d x} = 2 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 = 6

Find the line with slope m=

6

and passing through

(3, 1) : y = 6 x - 17

y = 6 x - 17

\int 1 + 4 x^{2} (8 x) d x

x \to \infty lim (- \frac{1}{0})

\frac{d}{d x} ((3 x^{2} + 1)^{2})

\int x (sec (2 y) + 2) d x

\frac{d}{d x} (\frac{1}{7})

What is the tangent of f(x)=x^2-8,\at x=3 ?
The tangent of f(x)=x^2-8,\at x=3 is y=6x-17