Use algebra to determine the location of the vertical asymptotes and holes in the graph of the function. (Enter your answers from smallest to largest.)
g(x)= 2x^2/x^4-x^2
Vertical asymptotes occur typically where the denominator equals zero.
So to determine these points, which are also holes on the graph, solve for the points where the denominator = 0, i.e.
x^4-x^2=0
x²(x²-1)=0
x=0 or ±x=1
To determine the location of the vertical asymptotes and holes in the graph of the function g(x), we need to analyze the denominator of the function, x^4 - x^2.
First, let's factor the denominator:
x^4 - x^2 = x^2(x^2 - 1)
To find the locations of the vertical asymptotes, we need to find the values of x that make the denominator equal to zero. Therefore, we set x^2 - 1 = 0 and solve for x:
x^2 - 1 = 0
(x + 1)(x - 1) = 0
From this, we find that x = -1 and x = 1 are the values of x that make the denominator zero.
Thus, x = -1 and x = 1 are the locations of the vertical asymptotes in the graph of the function g(x).
To find the location of any holes in the graph, we need to determine if there are any common factors between the numerator and the denominator that can be canceled out.
The numerator is 2x^2, and there are no factors in common with the denominator x^2(x^2 - 1).
Therefore, there are no holes in the graph of the function g(x).
In summary, the locations of the vertical asymptotes in the graph of the function g(x) are x = -1 and x = 1. There are no holes in the graph.
To determine the location of the vertical asymptotes and holes in the graph of the function, we need to analyze the factors in the denominator of the rational function.
First, let's factor the denominator, x^4 - x^2. We can factor out x^2 from both terms, giving us:
x^2(x^2 - 1)
Now, we have x^2 and (x^2 - 1) as factors in the denominator.
To find the vertical asymptotes, we look for the values of x that make the denominator equal to zero, since dividing by zero is undefined.
Setting x^2 - 1 equal to zero, we have:
x^2 - 1 = 0
We can solve this equation by factoring:
(x - 1)(x + 1) = 0
This gives us two possible values for x: x = 1 and x = -1.
Therefore, the vertical asymptotes of the function are x = 1 and x = -1.
To determine if there are any holes in the graph, we look for common factors between the numerator and denominator. In this case, we have a common factor of x^2.
So, if we cancel out the common factors, the simplified function would be:
g(x) = 2x^2 / (x^2 - 1)
Since there is no cancellation of factors that would result in a defined value at any particular x-value, there are no holes in the graph of the function.
In summary, the function g(x) = 2x^2 / (x^4 - x^2) has vertical asymptotes at x = 1 and x = -1, and no holes in its graph.