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.