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

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

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

tangent of

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

y = x - 6

Solution steps

Find the tangent point:

(2, - 4)

Find the slope of

f (x) = x^{2} - 3 x - 2 : \frac{df ( x )}{d x} = 2 x - 3

Compute the slope of

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

x = 2 : m = 1

Find the line with slope m=

1

and passing through

(2, - 4) : y = x - 6

y = x - 6

\frac{d}{d x} (- \frac{b}{x ^{2}})

\frac{d y}{d x} = \frac{cos ( 7 x )}{e ^{7 y}}, y (0) = 0

x \to 1 + lim (x^{2} - 2)

y^{^{''}} - 6 y + 13 y = 0

n = 1 \sum \infty 2 n - 1

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