Tangent LinesCalculus
posted by Greg .
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= 3x1 yet it's wrong. please explain thank you

y=(x^2+3) / (x+3)^{½}
y(1) = 4/2 = 2
y' = 3(x^2+4x1) / 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)
y2 = 3/4 (x1)
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