I think I have it, please check

  • Thread starter ashleyk
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In summary, the conversation discusses finding the derivative of a function and using it to find the tangent at a specific point. The derivative is found to be (-2x^2+10x-8)/(x^2-4)^2 and the tangent at point (0, f(0)) is calculated to be y=-.5x+5/4. The person asking for help confirms that the calculations are correct and compliments the method used.
  • #1
ashleyk
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Let f(x) be the function f(x)=(2x-5)/(x^2-4)

I found the derivative to be (-2x^2+10x-8)/(x^2-4)^2

I then had to find the tangent at point (0, f(0)) to be y=-.5x+5/4

Im not sure if it is right from the graph i made on my calculater. Any help would be great!
 
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  • #2
It's correct.
 
  • #3
ashleyk said:
Let f(x) be the function f(x)=(2x-5)/(x^2-4)
I found the derivative to be (-2x^2+10x-8)/(x^2-4)^2
I then had to find the tangent at point (0, f(0)) to be y=-.5x+5/4
Im not sure if it is right from the graph i made on my calculater. Any help would be great!

Yes,all the calculations are good.So,apparently,u have used a correct method,else u would have reached an incorrect result.

Keep up the good work!

Daniel.
 

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