What is the tangent of y=x^2+4 ?

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

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tangent of y=x^2+4

tangent of

y = 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

y = x^{2} + 4 : \frac{d y}{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} + 4) : y = 2 a_{0} x - a_{0}^{2} + 4

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

f (x^{3} - x)

\frac{\partial}{\partial y} (cos (y) + y e^{x})

\int 4 sec (4 x) tan (4 x) d x

x + 9, at x = 7

\frac{d}{d x} (\frac{2 x - 5}{( x ^{2} + 4 ) ^{3}})