How do you identify all asymptotes or holes for #f(x)=(x^3-7x^2+12x)/(-4x^2+8x)#?

Question

Serenity Jones · Accepted Answer

First step is to factor the cubic and quadratic equation on the numerator and denominator, respectively.

To factor the cubic expression you can either use a graphing calculator, or find a factor and carry on with synthetic or long division to find the other two factors.

If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0.

If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is y = a/b. A and B being the leading coefficients of the numerator and denominator.

If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Thus, in your equation, there will be no horizontal asymptote.

Ariana Alvarez · Answer

undefined To identify all asymptotes or holes for the function $ f(x) = \frac{x^3 - 7x^2 + 12x}{-4x^2 + 8x} $, we need to analyze its behavior near critical points and where the function is undefined.

First, we identify any vertical asymptotes by finding the values of $ x $ that make the denominator of the function zero, but not the numerator. These values represent the points where the function approaches infinity. So, for $ -4x^2 + 8x = 0 $, we find the critical points by setting the denominator equal to zero and solving for $ x $. This yields $ x = 0 $ and $ x = 2 $.

Next, we examine if any of these critical points are also zeros of the numerator. If they are, then they represent holes in the graph rather than vertical asymptotes. By substituting $ x = 0 $ and $ x = 2 $ into the numerator $ x^3 - 7x^2 + 12x $, we find that $ x = 0 $ is a zero of the numerator but $ x = 2 $ is not.

Therefore, $ x = 0 $ represents a hole in the graph.

In summary:
- Vertical asymptotes occur at $ x = 0 $ and $ x = 2 $.
- There is a hole in the graph at $ x = 0 $.