What is the tangent of f(x)=x^2+4 ?

The tangent of f(x)=x^2+4 is y=2a_{0}x-a_{0}^2+4

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tangent of f(x)=x^2+4

tangent of

f (x) = x^{2} + 4

y = 2 a_{0} x - a_{0}^{2} + 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 4)

Find the slope of

f (x) = x^{2} + 4 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 4

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

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} + 4) : y = 2 a_{0} x - a_{0}^{2} + 4

y = 2 a_{0} x - a_{0}^{2} + 4

\frac{d y}{d x} = \frac{1 + x y}{1 + x}

\frac{\partial}{\partial y} (x sin (x y))

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

e^{2 t s i n (2 t)}

\int - a cos (x) d x

What is the tangent of f(x)=x^2+4 ?
The tangent of f(x)=x^2+4 is y=2a_{0}x-a_{0}^2+4