Please help me solve a calc problem: find an equation of the tangent line to the graph of f(x)=(x+3)/(x^2+2) at x=1?

f(x) = (x + 3) / (x² + 2)

f'(x) = [(x² + 2)(x + 3)' - (x + 3)(x² + 2)'] / (x² + 2)²
= [(x² + 2) - (x + 3)(2x)] / (x² + 2)²
= (x² + 2 - 2x² - 6x) / (x² + 2)²
= (-x² -6x + 2) / (x² + 2)²

Slope of the tangent at (x = 1) = f'(1)
= [-1² -6(1) + 2] / (1² + 2)²
= -5/9

f(1) = (x + 3) / (x² + 2) = 4/3
Hence, (1, 4/3) is a point on the tangent.

Equation of the tangent of f(x) at x = 1 :
[y - (4/3)] / (x - 1) = -5/9
9 * [y - (4/3)] / (x - 1) = 9 * (-5/9)
(9y - 12) / (x - 1) = -5
9y - 12 = -5 * (x - 1)
9y - 12 = -5x + 5
5x + 9y = 17


The answer: 5x + 9y = 17

f(1) = 4/3 Point is (1,4/3)
f'(x) = [(x^2+2)-2x(x+3)]/(x^2+2)^2
f'(1) = [3-8]/9 = -5/9
y-4/3 = (-5/9)(x-1)
9y-12 = -5x+5
5x+9y = 17

Ans :

x=1,f(1)=4/3
f(x)=(x+3)/(x^2+2)--->f'(x)=-(x^2+6x-2)/(x^2+2)^2 -->f'(1)=-5/9
tangent y - (4/3)=(-5/9)(x-1)