What is the tangent of f(x)=x^2+9 ?

The tangent of f(x)=x^2+9 is y=2a_{0}x-a_{0}^2+9

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tangent of f(x)=x^2+9

tangent of

f (x) = x^{2} + 9

y = 2 a_{0} x - a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 9)

Find the slope of

f (x) = x^{2} + 9 : \frac{df ( x )}{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} + 9) : y = 2 a_{0} x - a_{0}^{2} + 9

y = 2 a_{0} x - a_{0}^{2} + 9

\int csc^{2} (2 t) d t

\int_{(- 1)}^{3} - ∣ x - 4 ∣ + 2 d x

x^{2} sin (\frac{1}{x})

\frac{x}{x ^{2} + 81}

\frac{\partial}{\partial y} (x e^{x^{3} y})

What is the tangent of f(x)=x^2+9 ?
The tangent of f(x)=x^2+9 is y=2a_{0}x-a_{0}^2+9