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

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

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

tangent of

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

y = 6 x - 15

Solution steps

Find the tangent point:

(3, 3)

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

x = 3 : m = 6

Find the line with slope m=

6

and passing through

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

y = 6 x - 15

6 e^{z}

\int \frac{1}{x ( x - 7 )} d x

\frac{d}{d y} (\frac{1}{1 + y ^{2}})

\frac{\partial}{\partial v} (u + uv)

y = x^{3} - 8 x^{2} + 4 x - 32

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