Determine the vertical asymptotes for the graph function:
f(x) = 4/x+10
g(x) = x^4 + 2/x
h(x) = x^2+4/x(x+5)(x-2)
f(x) = x+5/x^2 +4x - 32
I think you mean the limits
Limits? I don't get it?
I took algebra two and there not really verticle asymptotes threr limits you have been lied to which you should be used to if you are going to a public high school
All I remeber is you try to get them to cancel out and get removable discontinuties or something
this was a long time ago so I can't really help
To determine the vertical asymptotes for a function, we need to look for values of x that make the denominator equal to zero, but make the numerator non-zero.
Let's start with the first function:
f(x) = 4 / (x + 10)
Since the denominator is x + 10, the function has a vertical asymptote at x = -10. This means that as x approaches -10, the function approaches positive or negative infinity.
Now, let's analyze the second function:
g(x) = x^4 + 2 / x
In this case, the denominator is x. For a fraction to be undefined, the denominator must not equal zero. Therefore, there is no vertical asymptote for the function g(x) = x^4 + 2 / x.
Moving on to the third function:
h(x) = (x^2 + 4) / (x(x + 5)(x - 2))
Here, we must check for values of x that make the denominator equal to zero. The denominator is x(x + 5)(x - 2), so we set each factor equal to zero:
x = 0, x + 5 = 0, x - 2 = 0
Solving these equations, we find three values: x = 0, x = -5, and x = 2. Therefore, the function h(x) has vertical asymptotes at x = 0, x = -5, and x = 2.
Finally, let's consider the fourth function:
f(x) = (x + 5) / (x^2 + 4x - 32)
To determine the vertical asymptotes, we set the denominator equal to zero and solve for x:
x^2 + 4x - 32 = 0
Using the quadratic formula or factoring, we find two values for x: x = 4 and x = -8. Therefore, the function f(x) has vertical asymptotes at x = 4 and x = -8.