Math Problem Statement

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.

Solution

We are given the quadratic function y=x2+8x+15y = 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=ax2+bx+cy = ax^2 + bx + c. The sign of the coefficient aa determines whether the parabola opens upward or downward.

  • If a>0a > 0, the parabola opens upward.
  • If a<0a < 0, the parabola opens downward.

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

Part 2: Vertex of the Parabola

The vertex of a parabola in the form y=ax2+bx+cy = ax^2 + bx + c can be found using the formula for the x-coordinate of the vertex:

x=b2ax = \frac{-b}{2a}

For the given equation y=x2+8x+15y = x^2 + 8x + 15, we have a=1a = 1 and b=8b = 8. Plugging these values into the formula:

x=82(1)=82=4x = \frac{-8}{2(1)} = \frac{-8}{2} = -4

To find the y-coordinate of the vertex, substitute x=4x = -4 into the equation y=x2+8x+15y = x^2 + 8x + 15:

y=(4)2+8(4)+15=1632+15=1y = (-4)^2 + 8(-4) + 15 = 16 - 32 + 15 = -1

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

Part 3: Finding the x-intercepts

The x-intercepts occur when y=0y = 0. We solve the equation x2+8x+15=0x^2 + 8x + 15 = 0.

This is a quadratic equation, which can be factored:

x2+8x+15=(x+3)(x+5)=0x^2 + 8x + 15 = (x + 3)(x + 5) = 0

Setting each factor equal to zero:

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

Solving these equations gives:

x=3orx=5x = -3 \quad \text{or} \quad x = -5

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

Summary of Answers:

  • Part a: The parabola opens upward.
  • Part b: The vertex is (4,1)(-4, -1).
  • Part c: The x-intercepts are (3,0)(-3, 0) and (5,0)(-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=x2+8x+15y = 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 x2x^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!

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Parabolas

Formulas

Quadratic equation formula: y = ax^2 + bx + c
Vertex formula: x = -b / 2a
Factoring method for finding x-intercepts

Theorems

Properties of parabolas based on the sign of 'a'
Factoring quadratics

Suitable Grade Level

Grades 9-10