What is the tangent of f(x)=4x^2-8,\at 3 ?

The tangent of f(x)=4x^2-8,\at 3 is y=8a_{0}x-4a_{0}^2-8

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tangent of f(x)=4x^2-8,\at 3

tangent of

f (x) = 4 x^{2} - 8, at 3

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} - 8)

Find the slope of

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

Compute the slope of

f (x) = 4 x^{2} - 8

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

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

n = 1 \sum \infty \frac{3}{n}

\int \frac{2}{t - t ^{3}} d t

\frac{d y}{d x} = \frac{9 x + 2 y}{9 x + 2 y + 2}

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

What is the tangent of f(x)=4x^2-8,\at 3 ?
The tangent of f(x)=4x^2-8,\at 3 is y=8a_{0}x-4a_{0}^2-8