Explain when a trapezoid is not a parallelogram. Sketch an example.
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After applying <em>algebraic</em> analysis we find the <em>right</em> choices for each case, all of which cannot be presented herein due to <em>length</em> restrictions. Please read explanation below.
<h3>How to analyze rational functions</h3>
In this problem we have a rational function, whose features can be inferred by algebraic handling:
Holes - x-values that do not belong to the domain of the <em>rational</em> function:
x³ + 8 · x² - 9 · x = 0
x · (x² + 8 · x - 9) = 0
x · (x + 9) · (x - 1) = 0
x = 0 ∨ x = - 9 ∨ x = 1
But one root is an evitable discontinuity as:
y = (9 · x² + 81 · x)/(x³ + 8 · x² - 9 · x)
y = (9 · x + 81)/(x² + 8 · x - 9)
Thus, there are only two holes. (x = - 9 ∨ x = 1) Besides, there is no hole where the y-intercept should be.
Vertical asymptotes - There is a <em>vertical</em> asymptote where a hole exists. Hence, the function has two vertical asymptotes.
Horizontal asymptotes - <em>Horizontal</em> asymptote exists and represents the <em>end</em> behavior of the function if and only if the grade of the numerator is not greater than the grade of the denominator. If possible, this assymptote is found by this limit:
y = 0
The function has a horizontal asymptote.
x-Intercept - There is an x-intercept for all x-value such that numerator is equal to zero:
9 · x + 81 = 0
x = - 9
There is a x-intercept.
Lastly, we have the following conclusions:
- How many holes? 2
- One <em>horizontal</em> asymptote along the line where y always equals what number: 0
- This function has x-intercepts? True
- One <em>vertical</em> asymptote along the line where x always equals what number: 1
- There is a hole where the y-intercept should be? False
To learn more on rational functions: brainly.com/question/27914791
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