Determine the equation of the tangent line to the graph f(x)=12/(6-x^2), at the point (3,f(3)

1. Calculate \( f(3) \):
f(3) = \frac{12}{6 - 3^2} = \frac{12}{6 - 9} = \frac{12}{-3} = -4

2. Find the derivative \( f'(x) \) using the chain rule:
f(x) = \frac{12}{6 - x^2} = 12 \cdot (6 - x^2)^{-1}
f'(x) = 12 \cdot (-1) \cdot (6 - x^2)^{-2} \cdot (-2x)
f'(x) = \frac{24x}{(6 - x^2)^2}

3. Evaluate \( f'(x) \) at \( x = 3 \):
f'(3) = \frac{24 \cdot 3}{(6 - 3^2)^2} = \frac{72}{(6 - 9)^2} = \frac{72}{(-3)^2} = \frac{72}{9} = 8

4. Use the point-slope form of the equation of the line:
y - f(3) = f'(3)(x - 3)
y + 4 = 8(x - 3)
y + 4 = 8x - 24
y = 8x - 28

Final answer:
y = 8x - 28