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Sagot :
(a) (-∞, -8) U (-4, 0) U (5, 9)
To find the intervals of decreasing, you just pay attention to where the graph is angled downward, when looking at it from left to right.
I usually do this by drawing lines on the x value where there is a local max/min and looking to see what the values are at that line.
You then make an ordered pair out of the two numbers.
This is the same process for finding both the increasing and decreasing intervals.
(b) -4, 5
A common mistake for this problem could be that you put the maximum and minimum x values in for this function, but the problem is only asking for the maximums.
If you look at the graph and find the point, the x-values are what you will use for this problem.
(c) +
For these kinds of problems, I draw an imaginary vertical line through the middle of the function.
- If the far right and left parts of the function are going up, then the function is positive.
- If the far right and left parts of the function are going down, then the function is negative.
- If the far left part of the function is going down while the far right part of the function is going up, then the function is positive.
- If the far left part of the function is going up while the far right part of the function is going down, then the function is negative.
These positives/negatives belong to the leading coefficient of f.
(d) 6
The degree of f relies on the number of times the function touches the x-axis.
Multiplicity: the number of times any given x-intercept appears in the equation
- If an x-intercept passes through the x-axis, then that intercept has a multiplicity of 1
- If an x-intercept bounces off of the x-axis, then that intercept has a multiplicity of 2
- If an x-intercept passes through the x-axis, but flattens out when it touches the x-axis, then that intercept has a multiplicity of 3
Adding up the multiplicities of all of the x-intercepts gives you the degree of the function.
In this problem, there are x-intercepts that simply pass through the x-axis at -9, -6, 8, and 10, and an x-intercept that bounces off of the x-axis at 0, which makes it a function of degree 6.
(a) The function f is decreasing over the intervals (-∞, -8), (-4, 0), and (5, 7)
(b) The function f has local maximum at x = -4, and x = 5
(c) The sign of the leading coefficient is (positive) +
(d) The possibility for the degree of the polynomial is 6
The reasons for the above selected values are as follows:
(a) A function is decreasing over an interval where the values of the function decreases simultaneously as the input value increases
From the graph, in the region for the input x-values of -∞ to -8, we have;
The output y-value decreasing from infinity to -3
Therefore, the function decreases over the interval (-∞, -8)
Similarly, between the points (-4, 2), and (0, 0), we have:
The input x-values is increasing from x = -4 to x = 0
The output y-values is decreasing from y = 2 to y = 0
Therefore, the function is decreasing over the x-interval (-4, 0)
The output y-values of the function also decreases as the input values increase between the points (5, 4) and (7, -2), which is the interval (5, 7)
Therefore, the function decreases over the interval (5, 7)
The intervals over which the function decreases are;
(-∞, -8), (-4, 0), and (5, 7)
(b) A local maxima is a point that has a larger y-value, compared to other surrounding points
The point x is a local maximum if f(x) ≥ f(z) for all z in (a, b), where a < x < b
In the interval (-8, 0), f(-4) =2 > f(-8) = -3, and f(-4) = 2 > f(0) = 0
Therefore, the point (x, f(x)) = (-4, 2) is a local maximum point
Similarly, in the interval, (0, 9), we have; f(0) = 0 < f(5) = 4 > f(9) = -2 therefore, the point (x, f(x)) = (5, 4) is a local maxima
The local maxima are (-4, 2), and (5, 4), and the function f has local maximum at x = -4, and x = 5
(c) The leading coefficient is the numerical value that multiplies the variable having the highest exponent
The sign and number type of the leading coefficient determines the end behavior
[tex]\begin{array}{cll} \mathbf{Degree \ of \ the \ polynomial}&& \mathbf{Leading \ coefficient}\\\\&\mathbf{Positive (+)}&\mathbf{Negative(-)}\\\\Even&f(x) \rightarrow \infty \ as \ x \rightarrow \pm \infty&f(x) \rightarrow -\infty \ as \ x \rightarrow \pm \infty\\\\Odd &f(x) \rightarrow \infty \ as \ x \rightarrow \infty&f(x) \rightarrow \infty \ as \ x \rightarrow -\infty\\\\Odd&f(x) \rightarrow -\infty \ as \ x \rightarrow -\infty&f(x) \rightarrow -\infty \ as \ x \rightarrow \infty\end{array}[/tex]
The end behavior of the given graph is as x → ±∞, f(x) → ∞, therefore, the
sign of the leading coefficient is positive (+)
(d) The degree of the polynomial is given by the number and nature of the graph crossing of the x-axis
The graph crosses the x-axis and appears straight across the intercept at 4 points given 4 zeros
The graph bounces of the x-axis at (0, 0), giving 2 zeros
Therefore, the total number of zeros = 4 + 2 = 6 = The degree of the polynomial
The possible degree of the polynomial = 6
Learn more about polynomial function graphs here:
https://brainly.com/question/17155216
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