## University Calculus: Early Transcendentals (3rd Edition)

$f'(x) =1-9x^{-2}$ $slope = 0$ Equation of tangent line: $y = -6$
$f(x) = x +\frac{9}{x}$ $f(-3) = -3 -\frac{9}{3} = -6$ Now we know the point on the graph is: (-3, -6). So, $f'(x) = 1+9(-1)x^{-1-1}$ $f'(x) =1-9x^{-2}$ Therefore, the slope at x = -3 is: $slope = 0$ So, the equation of tangent is: $\frac{y+6}{x+3} = 0$ y+6 = 0 y = -6