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
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)