hello,

i have some question about rational functions.
1) are there any sites that are good in teaching graphing rational functions?

2) for the domain of the rational function it is what the denominator cannot equal to rite? for example the denominator of x^2-7x+10 can be factored to (x-2)(x-5) so the domain whould be X cannot equal to +2,+5 rite. these are also the Vertical asymptotes rite?

and last questions how would you expand x^3-9
thanks

wow

Hello! I'd be happy to answer your questions about rational functions.

1) There are several websites that are great for learning how to graph rational functions. Here are a few options you can explore:

- Khan Academy: Khan Academy has a comprehensive math section that covers various topics, including rational functions. They provide video tutorials and practice exercises on graphing rational functions.
- Mathway: Mathway is an online math problem solver that can help you graph rational functions. You can input your function, and it will generate a graph for you along with step-by-step explanations.
- Desmos: Desmos is an online graphing calculator that allows you to graph rational functions and explore their properties. It's user-friendly and provides a visual representation of the graph.

2) Yes, you're correct! In a rational function, the denominator cannot equal zero because division by zero is undefined. To find the domain of the function, you need to identify the values of x that make the denominator zero. In the case of the function x^2-7x+10, you correctly factored the denominator as (x-2)(x-5). To find the domain, you set each factor equal to zero and solve for x, which gives you x = 2 and x = 5. Therefore, the domain of the function is "x cannot equal 2 or 5," as you mentioned. These values also correspond to the vertical asymptotes of the graph.

As for your last question, to expand x^3-9, you can use the difference of cubes formula. The formula states that for any two numbers a and b, (a^3 - b^3) can be factored as (a - b)(a^2 + ab + b^2). In this case, a is x and b is 2√3. So, we have:

x^3 - 9 = (x)^3 - (2√3)^3
= (x - 2√3)(x^2 + (x)(2√3) + (2√3)^2)
= (x - 2√3)(x^2 + 2√3x + 12)

Therefore, x^3 - 9 is expanded as (x - 2√3)(x^2 + 2√3x + 12).

I hope this helps! Let me know if you have any more questions.