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

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

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tangent of f(x)=9x^2

tangent of

f (x) = 9 x^{2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

f (x) = 9 x^{2}

x = a_{0} : m = 18 a_{0}

Find the line with slope m=

18 a_{0}

and passing through

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

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

\int - 2 x^{3} + 12 x^{2} - 20 x + 8.5 d x

\frac{d y}{d x} + 7 x e^{y} = 0

\frac{d}{d x} (lo g_{4} (sec (x)))

\frac{d}{d x} (3 x^{2} - \frac{1}{x ^{2}})

16 y^{^{''}} + 8 y^{^{'}} + y = 0

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