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

The tangent of f(x)=4x^2 is y=8a_{0}x-4a_{0}^2

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tangent of f(x)=4x^2

tangent of

f (x) = 4 x^{2}

y = 8 a_{0} x - 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = 4 x^{2} : \frac{df ( x )}{d x} = 8 x

Compute the slope of

f (x) = 4 x^{2}

x = a_{0} : m = 8 a_{0}

Find the line with slope m=

8 a_{0}

and passing through

(a_{0}, 4 a_{0}^{2}) : y = 8 a_{0} x - 4 a_{0}^{2}

y = 8 a_{0} x - 4 a_{0}^{2}

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

\frac{d}{d x} (- 9 sin (x))

x \to - 1 lim (3 x - 2)

\frac{5}{x} (1.5)

What is the tangent of f(x)=4x^2 ?
The tangent of f(x)=4x^2 is y=8a_{0}x-4a_{0}^2