hi! The question is: a.)identify any horizontal and vertical asymptote and (b) identidify any holes in the graph. Verify your answers numerically by creating a table of values.
14.) f(x)= 3/(x-2)^3
16.) f(x)= x^2+2x+1)/2x^2-x-3
18.) 3-14x-5x^2)/3+7x+2x^2
I'm not looking for the answer necessarily, but I have no idea how to do this and the steps in my book aren't helping. Is there a way that whoever helps me can walk me through each step for each problem? Pretend you are teaching this to a 4th grader! Thank you!
if a denominator is zero, you have a vertical asymptote (look at 14 as x becomes 2)
if the function becomes constant you have a horizontal. look at 14 as x gets big positive and as x gets big negative
Ok that helps with the asymptotes... Could you help me with everything else?
holes occur where the numerator and denominator are both zero.
#16
x^2+2x+1 = (x+1)(x+1)
2x^2-x-3 = (x+1)(2x-3)
So, f(x) = (x+1)/(2x-3)
everywhere except at x = -1. At that point, f(x) = 0/0 which is not defined. The only vertical asymptote is at x = 3/2, and at x = -1 there is a hole. No matter how close you get to x = -1, f(x) gets very very close to 0, but at exactly x = -1, f(x) is undefined.
As x gets huge, f(x) is very close to x^2/2x^2 = 1/2, so y = 1/2 is the horizontal asymptote.
See the graph at
http://www.wolframalpha.com/input/?i=%28x^2%2B2x%2B1%29%2F%282x^2-x-3%29
Of course! I'm here to help and explain the steps to you in a simple and understandable way. Let's start with the first problem:
14.) f(x) = 3/(x-2)^3
(a) To identify the horizontal asymptote, we need to look at the behavior of the function as x approaches positive infinity and negative infinity.
As x approaches positive infinity (x → ∞), we can think of it as the denominator becoming very large. In this case, the denominator, (x-2)^3, will approach positive infinity because we are subtracting a positive number from a very large positive number. So, the function will approach zero or "hug" the x-axis as x gets larger.
As x approaches negative infinity (x → -∞), the behavior is similar. The denominator, (x-2)^3, will also approach positive infinity because we are subtracting a positive number from a very large negative number. Again, the function will approach zero or "hug" the x-axis as x gets smaller.
Therefore, the horizontal asymptote of f(x) = 3/(x-2)^3 is y = 0.
(b) To identify any holes in the graph, we need to check for any values of x that make the denominator zero, which would result in an undefined value for f(x). In this case, we see that if x = 2, the denominator (x-2)^3 would be zero and make the function undefined. So, there is a hole in the graph at x = 2.
To verify these answers numerically, we can create a table of values. Let's choose some x-values and evaluate f(x):
x | f(x)
-----------------
1.5 | -24
1.9 | -47.62
2 | undefined
2.1 | 47.62
3 | 12
We can see that as x approaches 2 from both sides (1.9 and 2.1), the function approaches a very large positive and negative value respectively. But at x = 2, the function is undefined.
I hope this explanation helps! Let's move on to the next problem.