Consider The Graph Of The Function F ( X ) = 2 ( X + 3 ) 2 + 2 F(x) = 2(x+3)^2 + 2 F ( X ) = 2 ( X + 3 ) 2 + 2 . Over Which Interval Is The Graph Decreasing?A. ( − ∞ , − 3 (-\infty, -3 ( − ∞ , − 3 ] B. ( − ∞ , 2 (-\infty, 2 ( − ∞ , 2 ] C. ( − 3 , ∞ (-3, \infty ( − 3 , ∞ ] D. ( 2 , ∞ (2, \infty ( 2 , ∞ ]
Introduction to Quadratic Functions
Quadratic functions are a fundamental concept in mathematics, and they play a crucial role in various fields, including algebra, geometry, and calculus. A quadratic function is a polynomial function of degree two, which means the highest power of the variable is two. The general form of a quadratic function is f(x) = ax^2 + bx + c, where a, b, and c are constants, and a cannot be equal to zero.
Graph of the Function f(x) = 2(x+3)^2 + 2
In this article, we will consider the graph of the function f(x) = 2(x+3)^2 + 2. To understand the graph of this function, we need to analyze its components. The function is a quadratic function with a leading coefficient of 2, which means it will open upwards. The vertex of the parabola is at (-3, 2), and the axis of symmetry is x = -3.
Finding the Interval of Decrease
To find the interval over which the graph of the function is decreasing, we need to find the intervals where the function is negative. Since the function is a quadratic function, it will have a maximum or minimum value at its vertex. In this case, the vertex is at (-3, 2), which is a maximum value.
Analyzing the Function
To analyze the function, we can start by finding the x-intercepts. The x-intercepts are the points where the function crosses the x-axis, and they can be found by setting the function equal to zero and solving for x.
Finding the x-Intercepts
To find the x-intercepts, we set the function equal to zero and solve for x:
2(x+3)^2 + 2 = 0
Subtracting 2 from both sides gives:
2(x+3)^2 = -2
Dividing both sides by 2 gives:
(x+3)^2 = -1
Since the square of any real number is non-negative, there is no real solution to this equation. This means that the function does not have any x-intercepts.
Finding the Intervals of Increase and Decrease
Since the function does not have any x-intercepts, we can conclude that the function is always positive. However, we need to find the intervals where the function is decreasing. To do this, we can use the first derivative of the function.
Finding the First Derivative
To find the first derivative of the function, we can use the power rule of differentiation:
f'(x) = d(2(x+3)^2 + 2)/dx
Using the power rule, we get:
f'(x) = 4(x+3)
Finding the Critical Points
To find the critical points, we set the first derivative equal to zero and solve for x:
4(x+3) = 0
Dividing both sides by 4 gives:
x+3 = 0
Subtracting 3 from both sides gives:
x = -3
Determining the Intervals of Increase and Decrease
To determine the intervals of increase and decrease, we can use the first derivative. If the first derivative is positive, the function is increasing. If the first derivative is negative, the function is decreasing.
Analyzing the First Derivative
To analyze the first derivative, we can start by finding the critical points. We already found the critical point at x = -3.
Determining the Sign of the First Derivative
To determine the sign of the first derivative, we can use a number line. We can choose a test point to the left of the critical point and a test point to the right of the critical point.
Choosing Test Points
Let's choose the test points x = -4 and x = -2.
Evaluating the First Derivative at the Test Points
Evaluating the first derivative at x = -4 gives:
f'(-4) = 4(-4+3) = -4
Evaluating the first derivative at x = -2 gives:
f'(-2) = 4(-2+3) = 4
Determining the Sign of the First Derivative
Since the first derivative is negative at x = -4 and positive at x = -2, we can conclude that the first derivative is negative to the left of the critical point and positive to the right of the critical point.
Determining the Intervals of Increase and Decrease
Since the first derivative is negative to the left of the critical point, the function is decreasing to the left of the critical point. Since the first derivative is positive to the right of the critical point, the function is increasing to the right of the critical point.
Conclusion
In conclusion, the graph of the function f(x) = 2(x+3)^2 + 2 is decreasing to the left of the critical point x = -3 and increasing to the right of the critical point x = -3.
Answer
The correct answer is C. (-3, ∞).
Final Thoughts
In this article, we analyzed the graph of the function f(x) = 2(x+3)^2 + 2 and determined the interval over which the graph is decreasing. We used the first derivative of the function to determine the intervals of increase and decrease. We also used a number line to analyze the sign of the first derivative. The correct answer is C. (-3, ∞).
Q: What is the vertex of the parabola?
A: The vertex of the parabola is at (-3, 2).
Q: What is the axis of symmetry?
A: The axis of symmetry is x = -3.
Q: Does the function have any x-intercepts?
A: No, the function does not have any x-intercepts.
Q: Is the function always positive?
A: Yes, the function is always positive.
Q: What is the first derivative of the function?
A: The first derivative of the function is f'(x) = 4(x+3).
Q: What is the critical point of the function?
A: The critical point of the function is x = -3.
Q: Is the function increasing or decreasing to the left of the critical point?
A: The function is decreasing to the left of the critical point.
Q: Is the function increasing or decreasing to the right of the critical point?
A: The function is increasing to the right of the critical point.
Q: What is the interval over which the graph of the function is decreasing?
A: The graph of the function is decreasing over the interval (-3, ∞).
Q: What is the correct answer?
A: The correct answer is C. (-3, ∞).
Q: Can you explain why the function is always positive?
A: Yes, the function is always positive because the leading coefficient is 2, which means the parabola opens upwards. The vertex is at (-3, 2), which is a maximum value. Since the parabola opens upwards, the function will always be positive.
Q: Can you explain why the function does not have any x-intercepts?
A: Yes, the function does not have any x-intercepts because the equation 2(x+3)^2 + 2 = 0 has no real solutions. This means that the function will never cross the x-axis.
Q: Can you explain why the first derivative is negative to the left of the critical point?
A: Yes, the first derivative is negative to the left of the critical point because the function is decreasing to the left of the critical point. This means that the function is decreasing as x decreases.
Q: Can you explain why the first derivative is positive to the right of the critical point?
A: Yes, the first derivative is positive to the right of the critical point because the function is increasing to the right of the critical point. This means that the function is increasing as x increases.
Q: Can you explain why the graph of the function is decreasing over the interval (-3, ∞)?
A: Yes, the graph of the function is decreasing over the interval (-3, ∞) because the first derivative is negative to the left of the critical point. This means that the function is decreasing as x decreases.
Q: Can you explain why the correct answer is C. (-3, ∞)?
A: Yes, the correct answer is C. (-3, ∞) because the graph of the function is decreasing over the interval (-3, ∞). This means that the function is decreasing as x decreases, and the interval (-3, ∞) is the correct interval over which the graph of the function is decreasing.