What is the tangent of f(x)=9x^2+5 ?

The tangent of f(x)=9x^2+5 is y=18a_{0}x-9a_{0}^2+5

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tangent of f(x)=9x^2+5

tangent of

f (x) = 9 x^{2} + 5

y = 18 a_{0} x - 9 a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 9 a_{0}^{2} + 5)

Find the slope of

f (x) = 9 x^{2} + 5 : \frac{df ( x )}{d x} = 18 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 = 18 a_{0}

Find the line with slope m=

18 a_{0}

and passing through

(a_{0}, 9 a_{0}^{2} + 5) : y = 18 a_{0} x - 9 a_{0}^{2} + 5

y = 18 a_{0} x - 9 a_{0}^{2} + 5

n = 0 \sum \infty \frac{1}{n ^{\frac{3}{2}}}

(x^{3} - 11 x^{2} + 30 x), (- x^{3} + 11 x^{2} - 30 x)

\frac{d}{d x} (lo g_{3} (\frac{x ^{2}}{x + 1}))

\frac{d y}{d x} = - 4 (y - 3)

\frac{\partial}{\partial x} (2 x + y^{6})

What is the tangent of f(x)=9x^2+5 ?
The tangent of f(x)=9x^2+5 is y=18a_{0}x-9a_{0}^2+5