What is the tangent of h(x)= 1/(x-2),(3,1) ?

The tangent of h(x)= 1/(x-2),(3,1) is y=-x+4

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tangent of h(x)= 1/(x-2),(3,1)

tangent of

h (x) = \frac{1}{x - 2}, (3, 1)

y = - x + 4

Solution steps

Find the slope of

h (x) = \frac{1}{x - 2} : \frac{d h ( x )}{d x} = - \frac{1}{( x - 2 ) ^{2}}

Compute the slope of

h (x) = \frac{1}{x - 2}

(3, 1) : m = - 1

Find the line with slope m=

- 1

and passing through

(3, 1) : y = - x + 4

y = - x + 4

\int \frac{2 x y}{( x ^{2} + y ^{2} ) ^{2}} d x

\int \frac{4}{x} d x

\frac{d ^{2}}{d x ^{2}} (f (x) g (x))

\int cos (2 x) sin (2 x) d x

\frac{d r}{d s} = 0.75 r

What is the tangent of h(x)= 1/(x-2),(3,1) ?
The tangent of h(x)= 1/(x-2),(3,1) is y=-x+4