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

Compute the slope of

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

x = a_{0} : 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

\frac{d y}{d t} = y (8 - y), y (0) = 9

\int sec^{2} (2 x) tan^{2} (2 x) d x

n = 0 \sum \infty \frac{9 ^{n}}{n !}

\frac{d}{d x} (3 x^{2} - 5)

y^{^{'}} = \frac{5 y}{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