SOLUTION: I am not sure I placed this question under the right category. It is h(x)=15x^2/x^2-1 find intercepts, asymtoses, and graph

Algebra ->  Rational-functions -> SOLUTION: I am not sure I placed this question under the right category. It is h(x)=15x^2/x^2-1 find intercepts, asymtoses, and graph      Log On


   



Question 549604: I am not sure I placed this question under the right category. It is
h(x)=15x^2/x^2-1 find intercepts, asymtoses, and graph

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I believe you meant h%28x%29=15x%5E2%2F%28x%5E2-1%29 That's what we call a rational function. Just like a rational number is the ratio of two integers, a rational function is the ratio of two polynomials, with polynomial defined loosely enough to include expressions like 2, or 5x, or 15x%5E2.
With rational functions you may often want to divide and factor polynomials to get equivalent expressions, like
h%28x%29=15x%5E2%2F%28%28x%2B1%29%28x-1%29%29
That form of the function makes it easy to see, what values of x make the function, zero, positive, negative, and undefined.
The numerator is always positive, except for x=0.
For x=0 the numerator and the function are zero.
The denominator is zero, and changes sign, at x=-1 and x=1. At those points, the function does not exist; it's undefined.
The denominator is positive for x>1, and for x<-1, and so is the function. The denominator and the function are negative for x values between -1, and 1.
As you approach those values of x (from the right or from the left), the numerator is pretty close to 15, but the denominator (negative or positive) approaches zero. So the function (negative or positive) increases in absolute value without limits. In calculus you would say that it tends to infinity, or some such thing. In pre-calculus, you just say that there are vertical asymptotes at x=-1 and x=1. Those are the (vertical) lines that the function "hugs".
You may not have learned about other asymptotes, but You have other asymptotes too.

The last expression lets you realize that, as x increases in absolute value (and so do x%5E2 and x%5E2-1), the function's value approaches 15. So the graph will hug the line y=15 at the far left and right. The line y=15 is a horizontal asymptote.
With all that information, you can graph th function, which should look like this:
graph%28300%2C300%2C-5%2C5%2C-15%2C35%2C15x%5E2%2F%28x%5E2-1%29%2C15%29%29 I would also draw the vertical lines representing the vertical asymptotes.