What is the tangent of f(x)=(x+2)/(x^2-9),\at x=1 ?

The tangent of f(x)=(x+2)/(x^2-9),\at x=1 is y=-7/32 x-5/32

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tangent of f(x)=(x+2)/(x^2-9),\at x=1

tangent of

f (x) = \frac{x + 2}{x ^{2} - 9}, at x = 1

y = - \frac{7}{32} x - \frac{5}{32}

Solution steps

Find the tangent point:

(1, - \frac{3}{8})

Find the slope of

f (x) = \frac{x + 2}{x ^{2} - 9} : \frac{df ( x )}{d x} = \frac{- x ^{2} - 4 x - 9}{( x ^{2} - 9 ) ^{2}}

Compute the slope of

f (x) = \frac{x + 2}{x ^{2} - 9}

x = 1 : m = - \frac{7}{32}

Find the line with slope m=

- \frac{7}{32}

and passing through

(1, - \frac{3}{8}) : y = - \frac{7}{32} x - \frac{5}{32}

y = - \frac{7}{32} x - \frac{5}{32}

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What is the tangent of f(x)=(x+2)/(x^2-9),\at x=1 ?
The tangent of f(x)=(x+2)/(x^2-9),\at x=1 is y=-7/32 x-5/32