Find the polynomial function (in simplest standard form) that would give a graph like the one shown. It crosses the x-axis at (-3, 0), (2, 0), and (5, 0). ful x

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.5: Graphs Of Functions
Problem 35E
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### Polynomial Function from a Graph

**Problem Statement:**
Find the polynomial function (in simplest standard form) that would give a graph like the one shown. It crosses the x-axis at (-3, 0), (2, 0), and (5, 0).

**Graph Description:**
The provided graph illustrates a polynomial curve that intersects the x-axis at three points: (-3, 0), (2, 0), and (5, 0). The y-axis is labeled with 'y' and the x-axis with 'x'. The polynomial appears to have a turning point or local extremum between the intercepts.

**Steps to Solution:**
Given the x-intercepts, the polynomial can be written in factored form as:
\[ P(x) = a(x + 3)(x - 2)(x - 5) \]

Here, \( a \) is a constant that can be any real number. To determine \( a \), additional information such as a point on the curve other than the intercepts would be required. In this case, if no such additional point is given, we can assume \( a = 1 \) for simplicity:

\[ P(x) = (x + 3)(x - 2)(x - 5) \]

To find the simplest standard form, expand the factors:

1. Multiply \( (x + 3) \) and \( (x - 2) \):
\[ (x + 3)(x - 2) = x^2 - 2x + 3x - 6 = x^2 + x - 6 \]

2. Multiply the result by \( (x - 5) \):
\[ (x^2 + x - 6)(x - 5) = x^3 - 5x^2 + x^2 - 5x - 6x + 30 \]

3. Combine like terms:
\[ x^3 - 4x^2 - 11x + 30 \]

Thus, the polynomial in simplest standard form is:
\[ P(x) = x^3 - 4x^2 - 11x + 30 \]

**Final Polynomial:**
\[ P(x) = x^3 - 4x^2 - 11x + 30 \]

This polynomial function satisfies the given x-intercepts as shown in the graph.
Transcribed Image Text:### Polynomial Function from a Graph **Problem Statement:** Find the polynomial function (in simplest standard form) that would give a graph like the one shown. It crosses the x-axis at (-3, 0), (2, 0), and (5, 0). **Graph Description:** The provided graph illustrates a polynomial curve that intersects the x-axis at three points: (-3, 0), (2, 0), and (5, 0). The y-axis is labeled with 'y' and the x-axis with 'x'. The polynomial appears to have a turning point or local extremum between the intercepts. **Steps to Solution:** Given the x-intercepts, the polynomial can be written in factored form as: \[ P(x) = a(x + 3)(x - 2)(x - 5) \] Here, \( a \) is a constant that can be any real number. To determine \( a \), additional information such as a point on the curve other than the intercepts would be required. In this case, if no such additional point is given, we can assume \( a = 1 \) for simplicity: \[ P(x) = (x + 3)(x - 2)(x - 5) \] To find the simplest standard form, expand the factors: 1. Multiply \( (x + 3) \) and \( (x - 2) \): \[ (x + 3)(x - 2) = x^2 - 2x + 3x - 6 = x^2 + x - 6 \] 2. Multiply the result by \( (x - 5) \): \[ (x^2 + x - 6)(x - 5) = x^3 - 5x^2 + x^2 - 5x - 6x + 30 \] 3. Combine like terms: \[ x^3 - 4x^2 - 11x + 30 \] Thus, the polynomial in simplest standard form is: \[ P(x) = x^3 - 4x^2 - 11x + 30 \] **Final Polynomial:** \[ P(x) = x^3 - 4x^2 - 11x + 30 \] This polynomial function satisfies the given x-intercepts as shown in the graph.
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