What is the tangent of f(x)=x^2-4,\at 5 ?

The tangent of f(x)=x^2-4,\at 5 is y=2a_{0}x-a_{0}^2-4

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tangent of f(x)=x^2-4,\at 5

tangent of

f (x) = x^{2} - 4, at 5

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

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

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

\int x sin (x) d x

ln (1)

x \to \infty lim (\frac{11}{2})

\int_{2}^{5} \frac{1}{x ^{2} - 1} d x

\frac{\partial}{\partial z} (e^{2 z} x + 2 z)

What is the tangent of f(x)=x^2-4,\at 5 ?
The tangent of f(x)=x^2-4,\at 5 is y=2a_{0}x-a_{0}^2-4