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^{^{'}} = - \frac{9 x}{16 y}

y^{^{''}} + y = 3 e^{- 8 x}, y (0) = 0, y^{^{'}} (0) = 0

x \to 4 lim (16 - x^{2})

x \to - 2 lim (5)

y = x^{3} - 4 x + 3

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