Find the horizontal asymptotes of the curve.
y=-6x/squareroot (4x^2-160
correction 4x^2-16
y = 6x/ 4(x^2-4)
undefined at x = +2 and x = -2 of course
y ---> 0 as x ---> +oo and as x ---> -oo
To find the horizontal asymptotes of the curve represented by the equation y = -6x / √(4x^2 - 160), we need to examine the behavior of the function as x approaches positive and negative infinity.
First, let's rewrite the equation in a more simplified form:
y = -6x / √(4x^2 - 160)
= -6x / (2√(x^2 - 40))
= -3x / √(x^2 - 40)
Now, as x approaches positive or negative infinity, we need to consider the term in the denominator: √(x^2 - 40).
Since x is becoming infinitely large, the term x^2 will dominate the expression x^2 - 40. Thus, the expression √(x^2 - 40) will effectively become √(x^2), which simplifies to |x| (the absolute value of x).
Now we can rewrite the equation as:
y = -3x / |x|
When x is positive, the equation becomes y = -3x / x = -3. Therefore, as x approaches positive infinity, the curve's y-value approaches -3.
When x is negative, the equation becomes y = -3x / -x = 3. Therefore, as x approaches negative infinity, the curve's y-value approaches 3.
Hence, the horizontal asymptotes of the curve are y = -3 and y = 3.