**Graphing a Quadratic Function****Objective:**Graph the quadratic function. Label the Vertex, Line of Symmetry (LOS), and intercepts.**Function:**\[ f(x) = -(x + 2)^2 + 5 \]**Steps:**1. **Identify the Vertex:** - The function is in vertex form \[ f(x) = a(x-h)^2 + k \]. - Here, $ a = -1 $, $ h = -2 $, $ k = 5 $. - Vertex: $(-2, 5)$2. **Line of Symmetry (LOS):** - The line of symmetry is the vertical line that passes through the vertex. - Equation: $ x = -2 $.3. **Intercepts:** - **y-intercept:** Set $ x = 0 $. \[ f(0) = -(0 + 2)^2 + 5 \] \[ f(0) = -4 + 5 = 1 \] - y-intercept: $(0, 1)$ - **x-intercepts:** Set $ f(x) = 0 $. \[ 0 = -(x + 2)^2 + 5 \] \[ -(x + 2)^2 = -5 \] \[ (x + 2)^2 = 5 \] \[ x + 2 = ±\sqrt{5} \] \[ x = -2 ± \sqrt{5} \]**Graph Description:**- The graph is a downward-opening parabola due to the negative coefficient ($ a = -1 $).- The vertex $(-2, 5)$ is the highest point on the graph.- The LOS is at $ x = -2 $.- The parabola crosses the y-axis at $ (0, 1) $.- The x-intercepts are located at $(-2 + \sqrt{5}, 0)$ and $(-2 - \sqrt{5}, 0)$.

Answered: Graph. Label Vertex, LOS, intercepts…

Math

Trigonometry

Graph. Label Vertex, LOS, intercepts f(x) = - (x + 2)² + 5

Trigonometry (11th Edition)

11th Edition

ISBN:9780134217437

Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Chapter1: Trigonometric Functions

Section: Chapter Questions

Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.

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Concept explainers

Maximization

Have you ever seen a bridge in the shape of a parabola? Where is its maximum on the bridge? The maximum of that bridge is the highest point on that bridge. The process of finding the maximum of the bridge here is called the maximization.

Question

**Graphing a Quadratic Function**

**Objective:**
Graph the quadratic function. Label the Vertex, Line of Symmetry (LOS), and intercepts.

**Function:**
\[ f(x) = -(x + 2)^2 + 5 \]

**Steps:**

1. **Identify the Vertex:**
- The function is in vertex form \[ f(x) = a(x-h)^2 + k \].
- Here, $ a = -1 $, $ h = -2 $, $ k = 5 $.
- Vertex: $(-2, 5)$

2. **Line of Symmetry (LOS):**
- The line of symmetry is the vertical line that passes through the vertex.
- Equation: $ x = -2 $.

3. **Intercepts:**

- **y-intercept:** Set $ x = 0 $.
\[ f(0) = -(0 + 2)^2 + 5 \]
\[ f(0) = -4 + 5 = 1 \]
- y-intercept: $(0, 1)$

- **x-intercepts:** Set $ f(x) = 0 $.
\[ 0 = -(x + 2)^2 + 5 \]
\[ -(x + 2)^2 = -5 \]
\[ (x + 2)^2 = 5 \]
\[ x + 2 = ±\sqrt{5} \]
\[ x = -2 ± \sqrt{5} \]

**Graph Description:**
- The graph is a downward-opening parabola due to the negative coefficient ($ a = -1 $).
- The vertex $(-2, 5)$ is the highest point on the graph.
- The LOS is at $ x = -2 $.
- The parabola crosses the y-axis at $ (0, 1) $.
- The x-intercepts are located at $(-2 + \sqrt{5}, 0)$ and $(-2 - \sqrt{5}, 0)$.

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