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

Question

write the equation of this line

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

Answer 1

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)