Points A (-10,-6) and B (6,2) are the endpoints of AB. What are the coordinates of point C on AB such that AC is 3/4 the length
1 answer:
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Answer:
- Domain: All the real values except x = 2 and x = 4: R - {2, 4}
- Holes: x = 2
- VA, vertical asymptores: x = 4
- HA: horizontal asymptotes: there are not horizontal asymptotes
- OA: oblique asymptotes: x + 6 [note that OH does not stand for any known feature, and so it is understood that it was intended to write OA]
- Roots: x = 2
- Y-intercept: -1
Step-by-step explanation:
1. <u>Given</u>:
- Note that the number 8 in the numerator is not part of the power.
- Type of function: rational function
2. <u>Domain</u>: is the set of x-values for which the function is defined.
The given function is defined for all x except those for which the denominator equals 0.
- Denominator: x² -6x + 8 = 0
- Solve for x:
Factor. (x - 4 )(x - 2) = 0
Zero product property: (x - 4) = 0 or (x - 2) = 0
x - 4 = 0 ⇒ x = 4
x - 2 = 0 ⇒ x = 2
- Domain:
All the real values except x = 2 and x = 4: x ∈ R / x ≠ 2 and x ≠ 4.
3. <u>Holes</u>:
The holes on the graph of a rational function are at those x-values for which both the numerator and denominator are zero.
- Find the values for which the numerator is zero:
Numerator: x³ - 8 = 0
Factor using difference of cubes property:
a³ - b³ = (a - b)(a² + ab + b²)
x³ - 8 = (x - 2)(x² + 2x + 4) = 0
Zero product property: (x - 2)(x² + 2x + 4) = 0
x - 2 = 0 ⇒ x = 2
x² + 2x + 4 = 0 (this has not real solution)
- The values for which the denominator is zero were determined above: x = 2 and x = 4.
- Conclusion: for x = 2 both numerator and denominator equal 0, so this is a hole.
4. <u>VA: Vertical asymptotes</u>.
The vertical asymptotes on the graph of a rational function are the vertical lines for which only the denominator (and not the numerator) equals zero.
- In the previous part it was determined that happens when x = 4.
5. <u>HA: Horizontal asymptotes</u>.
In rational functions, if the numerator is a higher degree polynomial than the denominator, there is no horizontal asymptote.
6. <u>OA: oblique asymptotes</u>
- Find the quotient and the remainder.
x + 6
_______________
x² - 6x + 8 ) x³ + 0x² + 0x - 8
- x³ + 6x² - 8x
___________
6 x² - 8x - 8
- 6x² + 36x - 48
_____________
28x - 56
Result: (x + 6) + (28x - 56) / (x² - 6x + 8)
- Find limit x → ∞
<u>7. Roots</u>:
Roots are the values for which f(x) = 0.
That happens when the numerator equals 0, and the denominator is not 0.
As determined earlier: x³ - 8 = 0 ⇒ x = 2.
8. <u>Y-Intercept</u>
The y-intercepts of any function are the y-values when x = 0
- Substitute x = 0 into the function:
Answer:
Linear function with rate of change/growth = 2.5, which agrees with the fourth statement listed in the answers options.
Step-by-step explanation:
Notice that both columns of reported x and y values are increasing.
Then examine how the given x-values increase:
-2, 2, 6, 10, 14 (in steps of 4 units)
and how their corresponding y-values increase:
-2, 8, 18, 28, 38 (in steps of 10 units)
therefore, if we do the rate of change for any pair and
, we get the following constant rate of change:
Given that this relationship is valid for any pair of (x,y) values. we conclude that the rate of increase is constant, and therefore we are in the presence of a linear function, whose rate of change is 2.5
Answer:
A, B, E
Step-by-step explanation:
The attachment shows a graph of the function. It tends to infinity for x going to infinity in either direction. (A, B are true)
The right-side piece is cubic, not quadratic.
The U-shape tells you the function is decreasing on the left side.
The two function definitions have the same value at x=2, so the function has no discontinuities. It is continuous. (E is true)
