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Can someone check if I did this right:

Find the equation of the tangent line to the graph of [tex]f(x) = \frac {2x-5}{x+1}[/tex] at the point at which x = 0.

[tex]f(x) = \frac {2(0) - 5}{(0) + 1} = -5 [/tex]

So I got y = -5

Then I wanted to find the derivative of the original using the quotient rule to find the slope.

[tex] f'(x) = \frac{(2x-5)(1) - (2)(x+1)}{(x+1)^2}

= \frac{2x-5 - 2x -2}{(x+1)^2} [/tex]

Is the slope -7? I plugged in 0 for x.

If so, I get :

y -y1 = m(x-x1)

y - -5 = m(x-0)

y + 5 = -7x + 0

y = -7x -5

Find the equation of the tangent line to the graph of [tex]f(x) = \frac {2x-5}{x+1}[/tex] at the point at which x = 0.

[tex]f(x) = \frac {2(0) - 5}{(0) + 1} = -5 [/tex]

So I got y = -5

Then I wanted to find the derivative of the original using the quotient rule to find the slope.

[tex] f'(x) = \frac{(2x-5)(1) - (2)(x+1)}{(x+1)^2}

= \frac{2x-5 - 2x -2}{(x+1)^2} [/tex]

Is the slope -7? I plugged in 0 for x.

If so, I get :

y -y1 = m(x-x1)

y - -5 = m(x-0)

y + 5 = -7x + 0

y = -7x -5

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