What is the tangent of y=(2x-5)/(x+1),\at x=0 ?

The tangent of y=(2x-5)/(x+1),\at x=0 is y=7x-5

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tangent of y=(2x-5)/(x+1),\at x=0

tangent of

y = \frac{2 x - 5}{x + 1}, at x = 0

y = 7 x - 5

Solution steps

Find the tangent point:

(0, - 5)

Find the slope of

y = \frac{2 x - 5}{x + 1} : \frac{d y}{d x} = \frac{7}{( x + 1 ) ^{2}}

Compute the slope of

y = \frac{2 x - 5}{x + 1}

x = 0 : m = 7

Find the line with slope m=

7

and passing through

(0, - 5) : y = 7 x - 5

y = 7 x - 5

(4, 43)

(- 3, 4)

(6, \frac{5 π}{4})

(- 2, - 2)

f = 1

What is the tangent of y=(2x-5)/(x+1),\at x=0 ?
The tangent of y=(2x-5)/(x+1),\at x=0 is y=7x-5