$$ y=x^{2}+12 x+20 $$ 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. b. Find the vertex. The vertex is $$ (-6,-16) $$. (Type an ordered pair.) c. Find the $$ x $$-intercepts. The $$ x $$-intercepts are (Type an integer or a fraction. Use a comma to separate answers as needed.)

Question

UpStudy AI · Accepted Answer

**a. Direction the Parabola Opens**

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

**b. Vertex of the Parabola**

The vertex $$(h, k)$$ of a parabola described by  
$$
y = ax^2 + bx + c
$$  
can be found using  
$$
h = -\frac{b}{2a} \quad \text{and} \quad k = f(h).
$$

For this equation, $$a = 1$$ and $$b = 12$$. So,  
$$
h = -\frac{12}{2(1)} = -6.
$$

Now, compute $$k$$ by substituting $$x = -6$$ into the equation:  
$$
k = (-6)^2 + 12(-6) + 20 = 36 - 72 + 20 = -16.
$$

Thus, the vertex is  
$$
(-6, -16).
$$

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

To find the $$x$$-intercepts, set $$y = 0$$:

$$
x^2 + 12x + 20 = 0.
$$

We factor the quadratic:

$$
x^2 + 12x + 20 = (x + 10)(x + 2) = 0.
$$

Setting each factor equal to zero gives:

$$
x + 10 = 0 \quad \Rightarrow \quad x = -10,
$$
$$
x + 2 = 0 \quad \Rightarrow \quad x = -2.
$$

Thus, the $$x$$-intercepts are  
$$
-10, -2.
$$
**a. The parabola opens upward.** **b. The vertex is $$(-6, -16)$$.** **c. The $$x$$-intercepts are $$-10$$ and $$-2$$.

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Algebra Questions