What is the tangent of f(x)=x^2-6,(3,3) ?

The tangent of f(x)=x^2-6,(3,3) is y=6x-15

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tangent of f(x)=x^2-6,(3,3)

tangent of

f (x) = x^{2} - 6, (3, 3)

y = 6 x - 15

Solution steps

Find the slope of

f (x) = x^{2} - 6 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 6

(3, 3) : m = 6

Find the line with slope m=

6

and passing through

(3, 3) : y = 6 x - 15

y = 6 x - 15

\frac{d}{d x} (x ln (x^{2} + 1))

\int \frac{e ^{x}}{( 1 + e ^{x} )} d x

\int (3 + x) d x

\int p (p + 2)^{6} d p

\frac{d}{d x} (\frac{1}{x ^{2}} - \frac{3}{x ^{4}})

What is the tangent of f(x)=x^2-6,(3,3) ?
The tangent of f(x)=x^2-6,(3,3) is y=6x-15