$$ y=x^{2}+12 x+20 $$ a. Determine if the parabola whose equation is given opens upward or downward. Choose the correct answer below. The parabola opens downward. The parabola opens upward. The vertex is the vertex. (Type an ordered pair.) c. Find the x-intercepts. The x-intercepts are $$ -10,-2 $$. (Type an integer or a fraction. Use a comma to separate answers as needed.) d. Find the y-intercept. The $$ y $$-intercept is (Type an integer or a fraction.)

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**a. Determining the Direction of Opening**

The coefficient of $$ x^2 $$ is $$ 1 $$ (a positive number), so the parabola opens upward.

**b. Finding the Vertex**

The vertex $$ (h, k) $$ can be found using
$$
h = -\frac{b}{2a}
$$
Here, $$ a = 1 $$ and $$ b = 12 $$, so
$$
h = -\frac{12}{2\cdot1} = -6.
$$
Now, substitute $$ x = -6 $$ into the equation to find $$ k $$:
$$
k = (-6)^2 + 12(-6) + 20 = 36 - 72 + 20 = -16.
$$
Thus, the vertex is $$ (-6, -16) $$.

**c. Finding the x-intercepts**

Set $$ y = 0 $$:
$$
x^2 + 12x + 20 = 0.
$$
Factor the quadratic:
$$
(x+10)(x+2)=0.
$$
Therefore,
$$
x = -10 \quad \text{or} \quad x = -2.
$$
The x-intercepts are $$ -10, -2 $$.

**d. Finding the y-intercept**

Set $$ x = 0 $$:
$$
y = 0^2 + 12(0) + 20 = 20.
$$
The y-intercept is $$ 20 $$.
**a. The parabola opens upward.** **b. The vertex is at $$ (-6, -16) $$.** **c. The x-intercepts are $$ -10 $$ and $$ -2 $$.** **d. The y-intercept is $$ 20 $$.**

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