Math Problem Statement

Question content area top left Part 1 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 select: (-4, -1) 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 select: -3, -5 negative 3 comma negative 5. ​(Type an integer or a fraction. Use a comma to separate answers as​ needed.) Your answer is correct.B. There are no​ x-intercepts. Part 4 d. Find the​ y-intercept. The​ y-intercept is    15. ​(Type an integer or a​ fraction.) Part 5 e. Use the results from parts ​(a)dash​(d) to graph the quadratic function. Use the graphing tool to graph the function.

Solution

Let's go through each part step by step for the given quadratic function:

The function is:

y=x2+8x+15y = x^2 + 8x + 15

Part (a): Determine if the parabola opens upward or downward.

  • The coefficient of x2x^2 is positive (it is 1), which means the parabola opens upward.

Answer: The parabola opens upward.


Part (b): Find the vertex.

To find the vertex of a quadratic function in standard form y=ax2+bx+cy = ax^2 + bx + c, use the formula for the x-coordinate of the vertex: xvertex=b2ax_{\text{vertex}} = \frac{-b}{2a} For y=x2+8x+15y = x^2 + 8x + 15, we have:

  • a=1a = 1
  • b=8b = 8

So, xvertex=82(1)=82=4x_{\text{vertex}} = \frac{-8}{2(1)} = \frac{-8}{2} = -4

Now, substitute x=4x = -4 into the original equation to find the y-coordinate of the vertex: y=(4)2+8(4)+15=1632+15=1y = (-4)^2 + 8(-4) + 15 = 16 - 32 + 15 = -1

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

Answer: The vertex is (4,1)(-4, -1).


Part (c): Find the x-intercepts.

The x-intercepts occur when y=0y = 0. Set the equation equal to 0: 0=x2+8x+150 = x^2 + 8x + 15 Now, solve the quadratic equation. We can factor it: 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 x=3orx=5x = -3 \quad \text{or} \quad x = -5

Thus, the x-intercepts are x=3x = -3 and x=5x = -5.

Answer: The x-intercepts are (3,0)(-3, 0) and (5,0)(-5, 0).


Part (d): Find the y-intercept.

The y-intercept occurs when x=0x = 0. Substitute x=0x = 0 into the equation: y=(0)2+8(0)+15=15y = (0)^2 + 8(0) + 15 = 15

Thus, the y-intercept is 1515.

Answer: The y-intercept is 1515.


Part (e): Use the results to graph the quadratic function.

  • Vertex: (4,1)(-4, -1)
  • x-intercepts: (3,0)(-3, 0) and (5,0)(-5, 0)
  • y-intercept: (0,15)(0, 15)
  • The parabola opens upward (since a>0a > 0).

With this information, you can plot these key points and sketch the curve of the parabola.


Would you like to see more details or need help with graphing the function? Here are a few additional questions for you:

  1. How do you factor a quadratic equation like x2+8x+15x^2 + 8x + 15?
  2. What other methods could you use to find the roots of a quadratic equation besides factoring?
  3. How does the sign of aa in the quadratic equation affect the direction of the parabola?
  4. Can you identify the axis of symmetry for the quadratic equation y=x2+8x+15y = x^2 + 8x + 15?
  5. What are the properties of the vertex in a parabola?

Tip: When finding the vertex, always remember that the formula x=b2ax = \frac{-b}{2a} gives you the axis of symmetry of the parabola!

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Vertex of a Parabola
X-Intercepts
Y-Intercept
Graphing Quadratics

Formulas

x_{vertex} = \frac{-b}{2a}
Quadratic equation formula ax^2 + bx + c = 0

Theorems

Factoring Quadratic Equations
Properties of Parabolas

Suitable Grade Level

Grades 9-10