The polynomial function of the least degree with integral coefficient that has the given zeros is <span>y = (x+2)(x-2)(x-4)(x-6). I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications about the said problem.</span>
Step-by-step explanation:
∫ (sec x − tan x) dx
∫ sec x dx + ∫ -tan x dx
∫ (sec²x + sec x tan x) / (sec x + tan x) dx + ∫ (-sin x / cos x) dx
ln(sec x + tan x) + ln(cos x) + C
Answer:
none above
Step-by-step explanation:
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
IT IS A HOPE THIS HELPS!!!
srry 4 caps