What is the tangent of f(x)=x^2-2 ?

The tangent of f(x)=x^2-2 is y=2a_{0}x-a_{0}^2-2

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tangent of f(x)=x^2-2

tangent of

f (x) = x^{2} - 2

y = 2 a_{0} x - a_{0}^{2} - 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 2)

Find the slope of

f (x) = x^{2} - 2 : \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 = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} - 2) : y = 2 a_{0} x - a_{0}^{2} - 2

y = 2 a_{0} x - a_{0}^{2} - 2

2 \frac{d y}{d x} = x y (x + 1)

\int x (2 x + 7)^{8} d x

\int \frac{1}{x ( 2 - x )} d x

\int \frac{1}{6 x - 1} d x

\int x^{2} e^{b x} d x

What is the tangent of f(x)=x^2-2 ?
The tangent of f(x)=x^2-2 is y=2a_{0}x-a_{0}^2-2