what is a function that Contains no vertical asymptotes but has a hole at x=2 and
another function that contains a horizontal asymptote of 1, vertical asymptotes of 2 and -3, and a hole at x=4.
For the first one, something like
y = (x-2)/(x^2 - 4) should do it
for the second, how about
y = x^2(x-4)/((x-2)(x+3)(x-4) )
thank you
no vertical asymptotes but has a hole at x=2
y = (x^2-2x)/(x-2)
To find a function that contains no vertical asymptotes but has a hole at x = 2, you can start by using the fact that vertical asymptotes occur when the denominator of a fraction becomes zero. Since we want to avoid vertical asymptotes, we need to make sure that the denominator does not become zero for any value of x.
One way to achieve this is by using a rational function with a numerator that has factors that cancel out those in the denominator. For example, consider the function:
f(x) = (x - 2) / (x - 2)
Here, we have a factor of (x - 2) in both the numerator and denominator, which cancels out. This eliminates the possibility of a vertical asymptote at x = 2. However, there is a hole at x = 2 because the function is not defined at that specific point.
Now, to find a function that contains a horizontal asymptote of 1, vertical asymptotes of 2 and -3, and a hole at x = 4, we can use similar concepts.
For the function to have a horizontal asymptote of 1, we need to make sure that the degree of the numerator is equal to or less than the degree of the denominator by dividing the leading terms.
To introduce vertical asymptotes at x = 2 and x = -3, we can have factors in the denominator that make those values zeros.
And finally, to have a hole at x = 4, we can cancel out common factors in the numerator and denominator.
One possible function that satisfies these conditions is:
g(x) = (x - 4) / (x^2 - x - 6)
This function has a vertical asymptote at x = 2 (since x^2 - x - 6 = (x - 3)(x + 2)), another vertical asymptote at x = -3, a hole at x = 4, and it approaches a horizontal asymptote of 1 as x approaches ±∞.