$$ y=x^{2}+12 x+20 $$ a. Determine if the parabola whose equation is given opens upward or downward. Choose the correct answer below. The parabola opens upward. The parabola opens downward. b. Find the vertex. The vertex is $$ (-6,-16) $$. (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. Direction of Opening**

The equation is  
$$
y = x^2 + 12x + 20
$$  
The coefficient of $$ x^2 $$ is $$ 1 $$, which is positive. Therefore, the parabola opens upward.

**b. Vertex**

For a quadratic $$ y = ax^2 + bx + c $$, the $$ x $$-coordinate of the vertex is  
$$
x = -\frac{b}{2a}.
$$  
Here, $$ a = 1 $$ and $$ b = 12 $$, so  
$$
x = -\frac{12}{2 \times 1} = -6.
$$  
Substitute $$ x = -6 $$ back into the equation to find the $$ y $$-coordinate:  
$$
y = (-6)^2 + 12(-6) + 20 = 36 - 72 + 20 = -16.
$$  
Thus, the vertex is  
$$
(-6, -16).
$$

**c. $$ x $$-Intercepts**

To find the $$ x $$-intercepts, set $$ y = 0 $$ and solve:  
$$
x^2 + 12x + 20 = 0.
$$  
We factor the quadratic:  
$$
x^2 + 12x + 20 = (x + 10)(x + 2) = 0.
$$  
Setting each factor equal to zero gives:  
$$
x + 10 = 0 \quad \Rightarrow \quad x = -10,
$$  
$$
x + 2 = 0 \quad \Rightarrow \quad x = -2.
$$  
So, the $$ x $$-intercepts are at $$ x = -10 $$ and $$ x = -2 $$.

**d. $$ y $$-Intercept**

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

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