Select The Part Of The Statement Where Hugh Made A Mistake.Hugh Wants To Graph The Function Below: $\[ G(x) = -x^3 + 3x^2 + 4x - 12 \\]He Made A Statement About The Characteristics Of The Graph Of Function \[$ G \$\].The Graph Of

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Introduction

Graphing functions is a fundamental concept in mathematics, and understanding the characteristics of a graph is crucial in various fields, including algebra, calculus, and engineering. In this article, we will analyze Hugh's statement about the characteristics of the graph of the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12. We will identify the part where Hugh made a mistake and provide a detailed explanation of the correct characteristics of the graph.

The Function

The given function is g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12. This is a cubic function, which means it has a degree of 3. The graph of a cubic function can have various characteristics, including a single maximum or minimum point, or no maximum or minimum point at all.

Hugh's Statement

Unfortunately, we do not have Hugh's exact statement about the characteristics of the graph of the function g(x)g(x). However, we can make some educated guesses based on common mistakes that students make when analyzing graphs.

Common Mistakes

When analyzing the graph of a function, students often make the following mistakes:

  • Incorrectly identifying the x-intercepts: Students may incorrectly identify the x-intercepts of the graph, which are the points where the graph crosses the x-axis.
  • Incorrectly identifying the y-intercept: Students may incorrectly identify the y-intercept of the graph, which is the point where the graph crosses the y-axis.
  • Incorrectly identifying the maximum or minimum point: Students may incorrectly identify the maximum or minimum point of the graph, which is the point where the graph changes from increasing to decreasing or vice versa.
  • Incorrectly identifying the intervals of increase and decrease: Students may incorrectly identify the intervals of increase and decrease of the graph, which are the intervals where the graph is increasing or decreasing.

Analyzing the Function

To analyze the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12, we need to find the x-intercepts, y-intercept, maximum or minimum point, and intervals of increase and decrease.

Finding the X-Intercepts

To find the x-intercepts, we need to set the function equal to zero and solve for x.

βˆ’x3+3x2+4xβˆ’12=0-x^3 + 3x^2 + 4x - 12 = 0

We can factor the left-hand side of the equation as follows:

βˆ’(xβˆ’4)(x2+4x+3)=0-(x - 4)(x^2 + 4x + 3) = 0

The first factor gives us the x-intercept x = 4. The second factor is a quadratic expression that can be factored as follows:

(x+1)(x+3)=0(x + 1)(x + 3) = 0

This gives us two more x-intercepts: x = -1 and x = -3.

Finding the Y-Intercept

To find the y-intercept, we need to evaluate the function at x = 0.

g(0)=βˆ’(0)3+3(0)2+4(0)βˆ’12=βˆ’12g(0) = -(0)^3 + 3(0)^2 + 4(0) - 12 = -12

So, the y-intercept is (0, -12).

Finding the Maximum or Minimum Point

To find the maximum or minimum point, we need to find the critical points of the function. The critical points are the points where the derivative of the function is equal to zero or undefined.

The derivative of the function is:

gβ€²(x)=βˆ’3x2+6x+4g'(x) = -3x^2 + 6x + 4

Setting the derivative equal to zero, we get:

βˆ’3x2+6x+4=0-3x^2 + 6x + 4 = 0

Solving this quadratic equation, we get:

x=βˆ’6Β±36βˆ’48βˆ’6x = \frac{-6 \pm \sqrt{36 - 48}}{-6}

x=βˆ’6Β±βˆ’12βˆ’6x = \frac{-6 \pm \sqrt{-12}}{-6}

This gives us two complex roots: x = 1 + 2i and x = 1 - 2i.

Since the critical points are complex, the function has no maximum or minimum point.

Finding the Intervals of Increase and Decrease

To find the intervals of increase and decrease, we need to evaluate the derivative of the function at various points.

The derivative of the function is:

gβ€²(x)=βˆ’3x2+6x+4g'(x) = -3x^2 + 6x + 4

Evaluating the derivative at x = 0, we get:

gβ€²(0)=βˆ’3(0)2+6(0)+4=4g'(0) = -3(0)^2 + 6(0) + 4 = 4

Since the derivative is positive, the function is increasing at x = 0.

Evaluating the derivative at x = 1, we get:

gβ€²(1)=βˆ’3(1)2+6(1)+4=7g'(1) = -3(1)^2 + 6(1) + 4 = 7

Since the derivative is positive, the function is increasing at x = 1.

Evaluating the derivative at x = 2, we get:

gβ€²(2)=βˆ’3(2)2+6(2)+4=βˆ’4g'(2) = -3(2)^2 + 6(2) + 4 = -4

Since the derivative is negative, the function is decreasing at x = 2.

Evaluating the derivative at x = 3, we get:

gβ€²(3)=βˆ’3(3)2+6(3)+4=βˆ’13g'(3) = -3(3)^2 + 6(3) + 4 = -13

Since the derivative is negative, the function is decreasing at x = 3.

Evaluating the derivative at x = 4, we get:

gβ€²(4)=βˆ’3(4)2+6(4)+4=βˆ’32g'(4) = -3(4)^2 + 6(4) + 4 = -32

Since the derivative is negative, the function is decreasing at x = 4.

So, the intervals of increase and decrease are:

  • The function is increasing on the interval (-∞, 1).
  • The function is decreasing on the interval (1, ∞).

Conclusion

In conclusion, Hugh made a mistake in his statement about the characteristics of the graph of the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12. The correct characteristics of the graph are:

  • The x-intercepts are x = 4, x = -1, and x = -3.
  • The y-intercept is (0, -12).
  • The function has no maximum or minimum point.
  • The intervals of increase and decrease are (-∞, 1) and (1, ∞).

Introduction

In our previous article, we analyzed the characteristics of the graph of the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12. We identified the x-intercepts, y-intercept, maximum or minimum point, and intervals of increase and decrease. In this article, we will answer some frequently asked questions about the graph of the function.

Q: What are the x-intercepts of the graph of the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12?

A: The x-intercepts of the graph of the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12 are x = 4, x = -1, and x = -3.

Q: What is the y-intercept of the graph of the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12?

A: The y-intercept of the graph of the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12 is (0, -12).

Q: Does the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12 have a maximum or minimum point?

A: No, the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12 has no maximum or minimum point.

Q: What are the intervals of increase and decrease of the graph of the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12?

A: The intervals of increase and decrease of the graph of the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12 are:

  • The function is increasing on the interval (-∞, 1).
  • The function is decreasing on the interval (1, ∞).

Q: How can I graph the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12?

A: You can graph the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12 using a graphing calculator or a computer algebra system. You can also use a graphing software or a programming language to graph the function.

Q: What are some real-world applications of the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12?

A: The function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12 has many real-world applications, including:

  • Modeling population growth or decline
  • Modeling the motion of an object under the influence of gravity
  • Modeling the behavior of a physical system, such as a spring-mass system

Conclusion

In conclusion, we have answered some frequently asked questions about the graph of the function g(x)=βˆ’x3+3x2+4xβˆ’12g(x) = -x^3 + 3x^2 + 4x - 12. We hope this article has been helpful in understanding the characteristics of the graph of the function. If you have any further questions or need further clarification, please feel free to ask.

Additional Resources

References