What is the tangent of f(x)=x^2-7,\at x=3 ?

The tangent of f(x)=x^2-7,\at x=3 is y=6x-16

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tangent of f(x)=x^2-7,\at x=3

tangent of

f (x) = x^{2} - 7, at x = 3

y = 6 x - 16

Solution steps

Find the tangent point:

(3, 2)

Find the slope of

f (x) = x^{2} - 7 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 7

x = 3 : m = 6

Find the line with slope m=

6

and passing through

(3, 2) : y = 6 x - 16

y = 6 x - 16

\frac{2}{x}, \frac{2}{x ^{2}}, 4

\frac{d}{d x} (4 arctan (x))

(y - x) e^{x} + (1 + e^{x}) y^{^{'}} = 0, y (1) = 1

\frac{d y}{d x}, 50 x^{\frac{1}{3}} y^{\frac{2}{3}} = 600

f (x) = (4 x)^{l n (4 x)}

What is the tangent of f(x)=x^2-7,\at x=3 ?
The tangent of f(x)=x^2-7,\at x=3 is y=6x-16