use long division to find the quotient:
x-2 I x² +6x -9
you don't have to finish, because the remainder doesn't matter. the answer is y=x+8
2)f(x) = 2/(x² + 3x - 10) = 2 / [(x + 5)(x - 2)]
Vertical asymptotes are x=2 and x=-5.
3)It's a linear function, just a line, with a slope of 4/7 and a y-intercept of 0 because b=0.the degree of the numerator is 1 and the degree of the denominator is 0. 1>0 so we do not have any horizontal asymptotes.
4) y = (7x+1)/(2x-9)
2xy - 9y = 7x + 1
2xy - 7x = 9y+1
x = (9y + 1) / (2y - 7)
6), y=x−6 is your oblique asymptote
7) line y=2x−3
8)There is a horizontal asymptote and two vertical asymptotes though.
They are y=0,x=13,andx=−2
9)The oblique or slant asymptote is 4x−5.
VA(the vertical asymptotes): x=-5, x=2
VA: x=3, x=6
There is no HA.
HA : y=7/2
There is no HA.
m = 1
b = -6
OA: y = x - 6
7. Same method:
m = 2
b = 0
OA: y = 2x
8. m = 0
There is no OA.
m = 4
1.Vertical asymptotes are at x=-5 and x=2 .
2.Vertical asymptotes are at x=6 and x=3.
3.No , horizontal asymptotes.
4.Horizontal asymptote .
5.No, horizontal asymptote.
6.Oblique asymptote at .
7.Oblique asymptote at y=2x-3
8.There is no oblique asymptote.
9.Oblique asymptote y=4x-5.
1. To find vertical asymptote
Substitute denominator is equal to zero
Factorize the polynomial we get
Hence, the vertical asymptote of f(x) at x= -5 and x=2
Fatorize the polynomial then we get
Therefore, the vertical asymptote of f(x) at x=6 and x=3
3.There is no horizontal asymptote of f(x) because the degree of numerator is greater than the degree of denominator.
4.Horizontal asymptote : The degree of numerator and degree of denominator are same therefore,
5.No horizontal asymptote because the degree of numerator is greater than the degree of denominator.
6.Oblique asymptote: oblique asymptote is the value of quotient which obtained by dividing the numerator by denominator .
Oblique asymptote of f(x) at y=x-6
7. Oblique asymptote of f(x) at y= 2x-3
8. There is no oblique asymptote because the degree of numerator is less than the degree of denominator.
9.Oblique asymptote of f(x) at y= 4x-5
C, The oblique asymptote for f(x) is steeper than for g(x).
The oblique asymptote fo f(x) is steeper than for g(x).
Just took the test and got it right