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

The tangent of f(x)=x^2-4,\at x=3 is y=6x-13

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

tangent of

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

y = 6 x - 13

Solution steps

Find the tangent point:

(3, 5)

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 = 3 : m = 6

Find the line with slope m=

6

and passing through

(3, 5) : y = 6 x - 13

y = 6 x - 13

\int e^{4 θ} sin (5 θ) d θ

f (x) = 3 x^{2}, g (x) = - 3 x + 6, [- 4, 2]

4 e^{x} - 3 x^{2}

x \to 48 lim (\frac{x ^{2}}{8} - \frac{6}{x})

\int x^{2} \cdot e^{4 x} d x

What is the tangent of f(x)=x^2-4,\at x=3 ?
The tangent of f(x)=x^2-4,\at x=3 is y=6x-13