How do I find the vertical and horizontal asymptotes of
f(x)= (x+2)/(5x^2)
Wouldn't one of them be (I,-I)?
How do I solve this?
Thank you!
as x gets big +, y ---> 0
as x gets big -, y----> -0
as x ----> -2 , y------> 0
as x ---> 0 , y ----> oo
GRAPH IT !!!!
To find the vertical and horizontal asymptotes of a function, you need to consider the behavior of the function as x approaches infinity or negative infinity.
First, let's determine the vertical asymptotes of f(x).
Vertical asymptotes occur when the denominator of the function approaches zero. So, we need to find the values of x that make the denominator, 5x^2, equal to zero.
To solve 5x^2 = 0, we set the equation equal to zero and solve for x:
5x^2 = 0
x^2 = 0
From this, we can see that x = 0.
Therefore, the vertical asymptote of f(x) is x = 0.
Next, let's consider the horizontal asymptote of f(x).
To find the horizontal asymptote, we need to compare the degrees of the numerator and denominator polynomials. In this case, the numerator, x + 2, is a linear polynomial, and the denominator, 5x^2, is a quadratic polynomial.
Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote of f(x) is y = 0.
So, the vertical asymptote is x = 0, and the horizontal asymptote is y = 0.
Regarding your suggestion of (I, -I), that notation represents a point in the complex plane (where I is the imaginary unit). However, for this rational function, there are no complex asymptotes; only vertical and horizontal asymptotes exist in the real number system.
I hope this explanation helps! Let me know if you have any further questions.