Asked by alex
f(x) = two divided by quantity x squared minus two x minus three
Graph
A coordinate axis scaled by one.
Domain and Range : _____________________
x and y Intercept(s) : _____________________
Horizontal Asymptote(s) : ___________________
Vertical Asymptote(s) : ____________________
f(x) = quantity x squared plus x minus two divided by quantity x squared minus three x minus four
Graph
A coordinate axis scaled by one.
Domain and Range : _____________________
x and y Intercept(s) : _____________________
Horizontal Asymptote(s) : ___________________
Vertical Asymptote(s): ___________________
Graph
A coordinate axis scaled by one.
Domain and Range : _____________________
x and y Intercept(s) : _____________________
Horizontal Asymptote(s) : ___________________
Vertical Asymptote(s) : ____________________
f(x) = quantity x squared plus x minus two divided by quantity x squared minus three x minus four
Graph
A coordinate axis scaled by one.
Domain and Range : _____________________
x and y Intercept(s) : _____________________
Horizontal Asymptote(s) : ___________________
Vertical Asymptote(s): ___________________
Answers
Answered by
Steve
Good grief. How about some math, instead of the noisy words? I'll do one.
f(x) = (x^2+x-2)/(x^2-3x-4)
All polynomials have domain of all reals.
Rationals have a domain of all reals except where the denominator is zero. Since
f(x) = (x+2)(x-1) / (x-4)(x+1)
The domain is all reals except x = -1 or 4
x-intercepts (where y=0) are at -2 and 1
y-intercept (where x=0) is at 1/2
Vertical asymptotes are where the denominator is zero and the numerator is not: x = -1 and 4. Note that those are where f(x) is not defined.
Horizontal asymptotes are where x is very large. In that case, f(x) is approximately
(x^2)/(x^2) = 1
Note that it is possible for the graph to cross the horizontal asymptotes, but at the extreme ends of the x-axis, the curve approaches them.
f(x) = (x^2+x-2)/(x^2-3x-4)
All polynomials have domain of all reals.
Rationals have a domain of all reals except where the denominator is zero. Since
f(x) = (x+2)(x-1) / (x-4)(x+1)
The domain is all reals except x = -1 or 4
x-intercepts (where y=0) are at -2 and 1
y-intercept (where x=0) is at 1/2
Vertical asymptotes are where the denominator is zero and the numerator is not: x = -1 and 4. Note that those are where f(x) is not defined.
Horizontal asymptotes are where x is very large. In that case, f(x) is approximately
(x^2)/(x^2) = 1
Note that it is possible for the graph to cross the horizontal asymptotes, but at the extreme ends of the x-axis, the curve approaches them.
Answered by
Henry
1. F(x) = 2/(x^2-2x-3)
C = -3 = 1*(-3). 1+(-3) = -2 = B.
F(x) = 2/(x+1)(x-3)
x+1 = 0
X = -1, Denominator = 0.
x-3 = 0
X = 3, Denominator = 0.
Domain = All real values of x except -1,
and 3.
Y-int. = 2/(0+1)(0-3) = 2/-3 = -2/3
No x-intercept, because the numerator is
constant and the fraction cannot be set
to 0.
2. F(x) = (x^2+x-2)/(x^2-3x-4)=
(x-1)(x+2)/(x+1)(x-4).
x+1 = 0
X = -1, Denominator = 0.
x-4 = 0
X = 4, Denominator = 0.
Domain = All real numbers except -1 and
4.
Y-int. = (0-1)(0+2)/(0+1)(0-4) = -2/-4 =
1/2.
Y = (x-1)(x+2)/(x+1)(x-4) = 0
(x-1)(x+2) = 0
x-1 = 0
X = 1 = x-int.
x+2 = 0
X = -2 = x-int.
X-Intercepts = 1, and -2.
C = -3 = 1*(-3). 1+(-3) = -2 = B.
F(x) = 2/(x+1)(x-3)
x+1 = 0
X = -1, Denominator = 0.
x-3 = 0
X = 3, Denominator = 0.
Domain = All real values of x except -1,
and 3.
Y-int. = 2/(0+1)(0-3) = 2/-3 = -2/3
No x-intercept, because the numerator is
constant and the fraction cannot be set
to 0.
2. F(x) = (x^2+x-2)/(x^2-3x-4)=
(x-1)(x+2)/(x+1)(x-4).
x+1 = 0
X = -1, Denominator = 0.
x-4 = 0
X = 4, Denominator = 0.
Domain = All real numbers except -1 and
4.
Y-int. = (0-1)(0+2)/(0+1)(0-4) = -2/-4 =
1/2.
Y = (x-1)(x+2)/(x+1)(x-4) = 0
(x-1)(x+2) = 0
x-1 = 0
X = 1 = x-int.
x+2 = 0
X = -2 = x-int.
X-Intercepts = 1, and -2.
Answered by
Jorgeeee
Thanks!
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