Find the vertical and horizontal asymptotes of the function
F(x)=x=(√9x^2)/(2x+1)
Not sure what the extra "x=" means, so I'll ignore it. And, assuming the usual carelssness with parentheses,
√(9x^2) = |3x|
so, f(x) = |3x|/(2x+1)
the vertical asymptotes occur when the denominator is zero. so where is
2x+1 = 0 ?
the horizontal asymptotes are at x = ±3/2
thanks
To find the vertical asymptotes of a rational function, we need to identify the values of x for which the denominator of the function becomes zero. In this case, the denominator is 2x+1. Setting the denominator equal to zero, we have:
2x + 1 = 0
Solving for x, we get:
2x = -1
x = -1/2
Therefore, there is a vertical asymptote at x = -1/2.
Next, let's find the horizontal asymptotes.
Horizontal asymptotes can be determined by examining the degrees of the numerator and denominator polynomials. In this case, the degree of the numerator is 1 (since x is the only term) and the degree of the denominator is also 1.
When the degrees of the numerator and denominator are the same, we can determine the horizontal asymptote by comparing the coefficients of the highest degree terms. In this case, the highest degree term in both the numerator and denominator is 9x^2.
Since the coefficients of the highest degree terms are 1 in both the numerator and denominator, the horizontal asymptote is given by the ratio of the leading coefficients. So, the horizontal asymptote is y = 9/2.
Therefore, the vertical asymptote is x = -1/2, and the horizontal asymptote is y = 9/2.