What is the tangent of y=(((x-3))/((x+1)))^2,\at x=0 ?

The tangent of y=(((x-3))/((x+1)))^2,\at x=0 is y=-24x+9

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tangent of y=(((x-3))/((x+1)))^2,\at x=0

tangent of

y = (\frac{( x - 3 )}{( x + 1 )})^{2}, at x = 0

y = - 24 x + 9

Solution steps

Find the tangent point:

(0, 9)

Find the slope of

y = (\frac{x - 3}{x + 1})^{2} : \frac{d y}{d x} = \frac{8 ( x - 3 )}{( x + 1 ) ^{3}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 24

Find the line with slope m=

- 24

and passing through

(0, 9) : y = - 24 x + 9

y = - 24 x + 9

\frac{x}{x + 1}

y^{^{''}} + 10 y = 0, y (0) = 2, y^{^{'}} (0) = 1

\int sin (4 x + π) d x

\frac{1}{7 x}

(sin (\frac{3}{x}))^{^{'}}

What is the tangent of y=(((x-3))/((x+1)))^2,\at x=0 ?
The tangent of y=(((x-3))/((x+1)))^2,\at x=0 is y=-24x+9