What are the $$ x $$-intercepts for $$ y=x^{2}+8 x+12 $$ $$ y $$

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1. To find the $$ x $$-intercepts, we set $$ y = 0 $$:
   $$
   x^2 + 8x + 12 = 0
   $$

2. Factor the quadratic. We need two numbers that multiply to $$ 12 $$ and add up to $$ 8 $$. The numbers $$ 2 $$ and $$ 6 $$ work since:
   $$
   2 \times 6 = 12 \quad \text{and} \quad 2 + 6 = 8
   $$
   Thus, we can factor the quadratic as:
   $$
   (x+2)(x+6) = 0
   $$

3. Set each factor equal to zero and solve for $$ x $$:
   $$
   x+2 = 0 \quad \Rightarrow \quad x = -2
   $$
   $$
   x+6 = 0 \quad \Rightarrow \quad x = -6
   $$

4. Therefore, the $$ x $$-intercepts are at:
   $$
   (-2,0) \quad \text{and} \quad (-6,0)
   $$
The $$ x $$-intercepts are at $$ (-2, 0) $$ and $$ (-6, 0) $$.

What are the -intercepts for

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