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Calculus

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An equation of the line tangent to the graph of y=(2x+3)/(3x-2) at the point (1,5) is...do I take the derivative of y=(2x+3)/(3x-2) to find the slope and then plug that into y-5=m(x-1)?

  • Calculus - ,

    yes

  • Calculus - ,

    I took the derivative and got -10/(3x-2)^2. So...I plug in 1 in the x. Then, I get the slope to be -10. Finally I plug it in (1,5) and m=-10 to get y=-10x+15. Is that correct?

  • Calculus - ,

    check your derivative
    I had y' = -13/(3x-2^2
    so the slope would be -13

  • Calculus - ,

    How??????

  • Calculus - ,

    I don't know. Why are you asking me?

  • Calculus - ,

    I believe the answer is y = -13x + 18. In the question I had, one of the choices was 13x + y = 18, which is what y = -13x + 18 equals. I got -13 for slope, then plugged it in to get y -5 = -13x + 13

  • Calculus - ,

    Y=-13+18

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