What is the tangent of f(x)=x^2-5 ?

The tangent of f(x)=x^2-5 is y=2a_{0}x-a_{0}^2-5

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tangent of f(x)=x^2-5

tangent of

f (x) = x^{2} - 5

y = 2 a_{0} x - a_{0}^{2} - 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 5)

Find the slope of

f (x) = x^{2} - 5 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 5

x = a_{0} : m = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} - 5) : y = 2 a_{0} x - a_{0}^{2} - 5

y = 2 a_{0} x - a_{0}^{2} - 5

y^{^{'''}} + 2 y^{^{''}} + 5 y^{^{'}} - 26 y = 0

y = \frac{x ^{2}}{2} + 4 x + 2, (- 3, - \frac{11}{2})

y^{^{'}} + \frac{1}{2} y = y^{3}

\frac{1}{2} e^{- t} cos (t) + \frac{1}{2} e^{- t} sin (t) - \frac{1}{2} e^{- 2 t}

\frac{d}{d x} (cos^{2} (ln (x)))

What is the tangent of f(x)=x^2-5 ?
The tangent of f(x)=x^2-5 is y=2a_{0}x-a_{0}^2-5