What is the tangent of x^2-9,\at x=2 ?

The tangent of x^2-9,\at x=2 is y=4x-13

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tangent of x^2-9,\at x=2

tangent of

x^{2} - 9, at x = 2

y = 4 x - 13

Solution steps

Find the tangent point:

(2, - 5)

Find the slope of

y = x^{2} - 9 : \frac{d y}{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 = 4

Find the line with slope m=

4

and passing through

(2, - 5) : y = 4 x - 13

y = 4 x - 13

f^{^{'}} (x) = 3

\frac{d}{d x} (1.1 + 18.9 x^{2})

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

\int \frac{( cos ( x ) - 1 )}{x} d x

\int \frac{1}{- \frac{1}{x ^{2}} + 2 x} d x