To determine the intervals on which the function is increasing, decreasing, concave up, or concave down, we need to analyze the first and second derivatives of the function.
1. Increasing Intervals:
The function is increasing when the first derivative, f'(x), is positive. Let's find the critical points by setting f'(x) = 0 and solving for x:
-(x + 2)ax^3 = 0
This equation gives us two critical points: x = 0 and x = -2.
Now let's analyze the sign of f'(x) in different intervals:
- For x < -2: Since f'(x) = -(x + 2)ax^3, and a > 0, f'(x) will be negative in this interval.
- For -2 < x < 0: f'(x) will be positive because we have a negative coefficient (-) multiplied by a positive value (a) and (x + 2) is negative in this interval.
- For x > 0: f'(x) will be positive because both the coefficient (-) and (x + 2) will be negative in this interval.
Therefore, the function is increasing on the interval (0, ∞).
2. Decreasing Intervals:
The function is decreasing when the first derivative, f'(x), is negative. From the analysis of increasing intervals, we can conclude that f'(x) is negative on the interval (-∞, -2).
Therefore, the function is decreasing on the interval (-∞, -2).
3. Concave Up Intervals:
The function is concave up when the second derivative, f''(x), is positive. Let's find the critical points by setting f''(x) = 0 and solving for x:
(2x + 6)ax^4 = 0
This equation gives us one critical point: x = -3.
Now let's analyze the sign of f''(x) in different intervals:
- For x < -3: Since f''(x) = (2x + 6)ax^4, and a > 0, f''(x) will be negative in this interval.
- For x > -3: f''(x) will be positive because both the coefficient (2) and (x + 3) will be positive in this interval.
Therefore, the function is concave up on the interval (-∞, -3).
4. Concave Down Intervals:
The function is concave down when the second derivative, f''(x), is negative. From the analysis of concave up intervals, we can conclude that f''(x) is negative on the interval (-3, ∞).
Therefore, the function is concave down on the interval (-3, ∞).
Now let's move on to sketching the graph of f(x) = 1 + ax + ax^2 and identifying the horizontal and vertical asymptotes.
To sketch the graph:
1. Determine the y-intercept: Substitute x = 0 into the equation f(x) = 1 + ax + ax^2 to find the y-intercept. The y-intercept is (0, 1).
2. Determine the x-intercepts: Set f(x) = 0 and solve for x. The equation ax^2 + ax + 1 = 0 can be solved using the quadratic formula.
3. Plot the y-intercept and x-intercepts on the graph.
4. Analyze the behavior of the graph as x approaches infinity (positive) and negative infinity. This will help identify the horizontal asymptotes.
5. Determine the critical points and inflection points. We have already found the critical points as x = 0, x = -2, and the inflection point as x = -3.
Regarding the vertical asymptotes, since the function is a polynomial, there are no vertical asymptotes.
The final step is to plot the graph of f(x) with the analyzed information. Unfortunately, as an AI text-based bot, I can't provide a visual representation. However, following the steps above will help you accurately sketch the graph and identify the asymptotes.