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March 25, 2017

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a line is tangent to the curve y= (x^2 + 3)/ (x+3)^1/2 at the point where x=1.

write the equation of this line

i get y= 3x-1 yet it's wrong. please explain thank you

  • Tangent Lines-Calculus - ,

    y=(x^2+3) / (x+3)½
    y(1) = 4/2 = 2

    y' = 3(x^2+4x-1) / 2(x+3)3/2
    y'(1) = 3(4)/2*8 = 3/4

    so, now you have a point (1,2) and a slope (3/4)

    y-2 = 3/4 (x-1)

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