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): ___________________
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.
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.
Thanks!
To find the graph, domain, range, x and y intercepts, horizontal asymptotes, and vertical asymptotes for the given functions, we can follow these steps:
1. Graphing:
To graph the function, we can plot several points on the coordinate plane. We will choose different x-values, calculate the corresponding y-values using the function, and plot the points.
2. Domain and Range:
The domain of a function is the set of all possible x-values for which the function is defined. The range is the set of all possible y-values that the function can produce.
3. X and Y intercepts:
The x-intercepts are the points where the graph of the function intersects the x-axis. They are found by setting y = 0 and solving for x. The y-intercepts are the points where the graph intersects the y-axis. They are found by setting x = 0 and solving for y.
4. Horizontal Asymptotes:
Horizontal asymptotes are horizontal lines that the graph of a function approaches as x approaches positive or negative infinity. They can be found by analyzing the behavior of the function as x approaches infinity or negative infinity.
5. Vertical Asymptotes:
Vertical asymptotes are vertical lines that the graph of a function approaches as x approaches a certain value. They can be found by analyzing the behavior of the function as x approaches a particular value.
Now, let's apply these steps to the given functions:
1. f(x) = 2 / (x^2 - 2x - 3)
To graph f(x), we can choose several x-values and calculate the corresponding y-values:
For example:
When x = -3, y = 2 / ((-3)^2 - 2(-3) - 3) = 2 / (9 + 6 - 3) = 2 / 12 = 1/6
When x = -2, y = 2 / ((-2)^2 - 2(-2) - 3) = 2 / (4 + 4 - 3) = 2 / 5
When x = -1, y = 2 / ((-1)^2 - 2(-1) - 3) = 2 / (1 + 2 - 3) = -2
When x = 0, y = 2 / (0^2 - 2(0) - 3) = 2 / (0 + 0 - 3) = -2/3
When x = 1, y = 2 / (1^2 - 2(1) - 3) = 2 / (1 - 2 - 3) = 2
When x = 2, y = 2 / (2^2 - 2(2) - 3) = 2 / (4 - 4 - 3) = -2
When x = 3, y = 2 / (3^2 - 2(3) - 3) = 2 / (9 - 6 - 3) = 2 / 0 (undefined)
Plotting these points on a coordinate plane, we can see the shape of the graph. Join the points with a smooth curve to get the graph of f(x).
The domain of f(x) is all real numbers except the values that make the denominator zero. In this case, we have x^2 - 2x - 3 = 0. Solving this quadratic equation, we find x = -1 and x = 3. So, the domain of f(x) is (-∞, -1) U (-1, 3) U (3, ∞).
The range of f(x) is all real numbers except for the values that make the function undefined. Since the range of f(x) corresponds to the y-values of the graph, we observe that the graph extends from negative infinity to positive infinity, excluding the point where the y-value is 0.
The x-intercepts are the points where the graph crosses the x-axis. From the graph, we can see that it intersects the x-axis at x = -1 and x = 3.
Since the y-intercept is the point where the graph crosses the y-axis, we can find it by setting x = 0 in the function: f(0) = 2 / (0^2 - 2(0) - 3) = 2 / (-3) = -2/3. So, the y-intercept is (0, -2/3).
As x approaches positive or negative infinity, the function approaches 0. Thus, the horizontal asymptote is y = 0.
There are no vertical asymptotes in this case.
2. f(x) = (x^2 + x - 2) / (x^2 - 3x - 4)
To find the graph, domain, range, x and y intercepts, horizontal asymptotes, and vertical asymptotes for this function, we can follow the same steps as above.
However, since the process can be quite lengthy to explain, I will provide you with the results directly:
-The graph of f(x) has a "U" shape, crossing the x-axis at x = -2 and x = 1, and touching the x-axis at x = 2.
-The domain of f(x) is all real numbers except x = -1 and x = 4, as these values make the denominator zero.
-The range of f(x) is all real numbers except the values that make the function undefined.
-The x-intercepts are (-2, 0) and (1, 0).
-The y-intercept is (0, -2).
-As x approaches positive or negative infinity, the function approaches 1.
-The vertical asymptotes are x = -1 and x = 4.
I hope this helps! Let me know if you have any further questions.