State the x-coordinate of any holes in the function.

Question

f(x)=(x+3)(6-x)

        ------------

        (x-2)(x+3)

 

A. -3

B. 3

C. 2

D. 0

E. None of the above

Answer 1

Hi, Terah.

 

Holes and asymptotes come from "problems" with the fraction.

You know that you cannot divide by zero, but what if you find an "x" that makes the denominator (bottom) zero?

 

These x-values are forbidden and this is graphed as either a hole (if you can cancel it out or "get rid of it") or an asymptote (an invisible vertical line that no graph can cross)

 

In your problem, there is an (x+3) in the top and the bottom so it will cancel out (go away).  The (x-2) will no go away, it is stuck in the denominator forever.

 

So if x=-3, the original denom would be zero (PROBLEM!), but it will cancel and so just a hole is let in the graph.

 

If x=2, there will also be a zero in the denom, but this one will not go away, so your graph will have a vertical asymptote at x=2.

 

So to answer your question....  There will be a hole at = -3

 

Do you have a graphing calculator?  If so, graph this function on your calculator and zoom in around x= -3. (Set view window x's to -3.05 min and -2.95 max,  y's to between -3 and 3).   You can actually see the hole in the graph!  It is really pretty amazing!  :-)