Pls help i will give brainliest
2 answers:
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First of all get a picture of what this looks like. Desmos is a pretty good graphing program, but anything that will do polar coordinates will work.
Here is the graph.
What you can see is that this graph is symmetrical around the x axis.
When you talk about symmetry, you can think of it as taking a mirror and putting it where you think there is symmetry. If you can't tell the difference between the image and the real thing, then you have symmetry.
In this case, the mirror will show symmetry along the x axis and no where else.
Here is the graph.
What you can see is that this graph is symmetrical around the x axis.
When you talk about symmetry, you can think of it as taking a mirror and putting it where you think there is symmetry. If you can't tell the difference between the image and the real thing, then you have symmetry.
In this case, the mirror will show symmetry along the x axis and no where else.
Answers:
Vertical asymptote: x = 0
Horizontal asymptote: None
Slant asymptote: (1/3)x - 4
<u>Explanation:</u>
d(x) =
=
Discontinuities: (terms that cancel out from numerator and denominator):
Nothing cancels so there are NO discontinuities.
Vertical asymptote (denominator cannot equal zero):
3x ≠ 0
<u>÷3</u> <u>÷3 </u>
x ≠ 0
So asymptote is to be drawn at x = 0
Horizontal asymptote (evaluate degree of numerator and denominator):
degree of numerator (2) > degree of denominator (1)
so there is NO horizontal asymptote but slant (oblique) must be calculated.
Slant (Oblique) Asymptote (divide numerator by denominator):
- <u>(1/3)x - 4 </u>
- 3x) x² - 12x + 20
- <u>x² </u>
- -12x
- <u>-12x </u>
- 20 (stop! because there is no "x")
So, slant asymptote is to be drawn at (1/3)x - 4