Equation of Tangent Line

  • Thread starter swears
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  • #1
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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
 
Last edited:

Answers and Replies

  • #2
StatusX
Homework Helper
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I think you got the quotient rule wrong. It's low dee high minus high dee low.
 
  • #3
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Not sure what you mean. I was told you can do it either of these two ways:

[tex]F'(\frac{f}{g}) = \frac{fg' - f'g}{g^2} or \frac {f'g - fg'}{g^2} [/tex]
 
Last edited:
  • #4
StatusX
Homework Helper
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They can't both be right: one is the negative of the other. You want the second one.
 
  • #5
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Ok so now I get: [tex] \frac{2x+1 - 2x+5}{(x+1)^2}[/tex]

Do I plug x = 0 into that to find my slope? Which would give me m = 6
 
  • #6
StatusX
Homework Helper
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Yes, everything else you did in the first post looks right.
 

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