Consider the following function. Complete parts​ (a) through​ (e) below.
y equals x squared plus 8 x plus 15
Question content area bottom left
Part 1
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.
Your answer is correct.
Part 2
b. Find the vertex.
The vertex is 
  
left parenthesis negative 4 comma negative 1 right parenthesis.
​(Type an ordered​ pair.)
Part 3
c. Find the​ x-intercepts. Select the correct answer​ and, if​ necessary, fill in the answer box to complete your choice. 
A.
The​ x-intercept(s) is/are 
  
enter your response here.
​(Type an integer or a fraction. Use a comma to separate answers as​ needed.)
B.
There are no​ x-intercepts. 

We are given the quadratic function \( y = x^2 + 8x + 15 \). Let's go through the different parts step by step:

### Part 1: Opening of the Parabola
The general form of a quadratic equation is \( y = ax^2 + bx + c \). The sign of the coefficient \( a \) determines whether the parabola opens upward or downward.

- If \( a > 0 \), the parabola opens **upward**.
- If \( a < 0 \), the parabola opens **downward**.

In this case, \( a = 1 \), which is positive. So, **the parabola opens upward**.

### Part 2: Vertex of the Parabola
The vertex of a parabola in the form \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:

\[
x = \frac{-b}{2a}
\]

For the given equation \( y = x^2 + 8x + 15 \), we have \( a = 1 \) and \( b = 8 \). Plugging these values into the formula:

\[
x = \frac{-8}{2(1)} = \frac{-8}{2} = -4
\]

To find the y-coordinate of the vertex, substitute \( x = -4 \) into the equation \( y = x^2 + 8x + 15 \):

\[
y = (-4)^2 + 8(-4) + 15 = 16 - 32 + 15 = -1
\]

Thus, the vertex is \( (-4, -1) \).

### Part 3: Finding the x-intercepts
The x-intercepts occur when \( y = 0 \). We solve the equation \( x^2 + 8x + 15 = 0 \).

This is a quadratic equation, which can be factored:

\[
x^2 + 8x + 15 = (x + 3)(x + 5) = 0
\]

Setting each factor equal to zero:

\[
x + 3 = 0 \quad \text{or} \quad x + 5 = 0
\]

Solving these equations gives:

\[
x = -3 \quad \text{or} \quad x = -5
\]

Thus, the x-intercepts are \( x = -3 \) and \( x = -5 \), or as an ordered pair: \( (-3, 0) \) and \( (-5, 0) \).

### Summary of Answers:
- **Part a**: The parabola opens **upward**.
- **Part b**: The vertex is \( (-4, -1) \).
- **Part c**: The x-intercepts are \( (-3, 0) \) and \( (-5, 0) \).

---

Would you like more details on any part of this process or have any other questions?

Here are 5 related questions for further practice:
1. What is the y-intercept of the function \( y = x^2 + 8x + 15 \)?
2. How do you find the axis of symmetry for a parabola?
3. Can a quadratic function have more than two x-intercepts?
4. How would the graph of the function change if the coefficient of \( x^2 \) were negative?
5. How do you complete the square to rewrite the function in vertex form?

**Tip**: Factoring is a quick way to find x-intercepts when the quadratic is easily factorable, but using the quadratic formula works for all cases!

Consider the following function. Complete parts (a) through (e) below. y = x^2 + 8x + 15. a. Determine if the parabola opens upward or downward. b. Find the vertex. c. Find the x-intercepts.

Learn how to solve the quadratic equation y = x^2 + 8x + 15 by determining the parabola's direction, finding the vertex, and calculating the x-intercepts. This problem covers key algebraic concepts and is ideal for students in Grades 9-10.

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