SOLUTION: My math question reads: Find the hole in this function {{{y= (3x-48)/(2x^2-29x-48)}}} Can you please help me figure this out?

Algebra ->  Algebra  -> Square-cubic-other-roots -> SOLUTION: My math question reads: Find the hole in this function {{{y= (3x-48)/(2x^2-29x-48)}}} Can you please help me figure this out?      Log On

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Question 176275: My math question reads:
Find the hole in this function y=+%283x-48%29%2F%282x%5E2-29x-48%29

Can you please help me figure this out?
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(48568) About Me  (Show Source):
You can put this solution on YOUR website!
Find the hole in this function y= (3x-48)/(2x^2-29x-38)
-factor to get:
----------------------------
y = [3(x-16)]/[2x^2-29x-38]
This function does not have a "hole".
--------------
An example of a function with a hole is f(x) = [(x-1)(x+2)]/[(x-1)]
This function has a hole at x = 1
====================
Cheers,
Stan H.

Answer by Edwin McCravy(6938) About Me  (Show Source):
You can put this solution on YOUR website!
Edwin's solution: I assumed the 38 was a typo which should have been 48, for then the function has a hole.

y=+%283x-48%29%2F%282x%5E2-29x-48%29

Factor the numerator and denominator:

y=+%283%28x-16%29%29%2F%28%28x-16%29%282x%2B3%29%29

DO NOT cancel the %28x-16%29's YET, for if
you do you will get a different function
that is like this function except the hole
at x=16 will be filled up:

You cannot substitute 16 into the above
function since it will cause the denominator
to be 0, due to the %28x-16%29.

The graph of that function is this.  See the hole 
where x is 16?  That's because you can't substitute
16 for x without getting 0 in the denominator.   

drawing%28400%2C400%2C-4%2C20%2C-2%2C2%2C%0D%0Agraph%28400%2C400%2C-4%2C20%2C-2%2C2%2C%28sqrt%28x-16.1%29%2Fsqrt%28x-16.1%29%29%283%28x-16%29%29%2F%28%28x-16%29%282x%2B3%29%29++++%29%2C%0D%0Agraph%28400%2C400%2C-4%2C20%2C-2%2C2%2C%28sqrt%2815.8-x%29%2Fsqrt%2815.8-x%29%29%283%28x-16%29%29%2F%28%28x-16%29%282x%2B3%29%29++++%29%2C%0D%0A%0D%0A%0D%0Alocate%2815.8%2C.18%2Co%29+%29

Now if you cancel the %28x-16%29's, you
will get a NEW function:
y=+%283%28x-16%29%29%2F%28%28x-16%29%282x%2B3%29%29

y=+%283%28cross%28x-16%29%29%29%2F%28%28cross%28x-16%29%29%282x%2B3%29%29

y=3%2F%282x%2B3%29

which is exactly like your given function except
the hole will be plugged up.  That's because with
the %28x-16%29's gone you can now substitute x=16 
into this new function.  In fact when you substitute 
x=16 into this new function, you get:

y=3%2F%282%2816%29%2B3%29
y=3%2F%2832%2B3%29
y=3%2F35

So the hole in the previous graph is the point (16,3%2F35).

The graph of the new function y=3%2F%282x%2B3%29 
is exactly like your original function except you will notice
that the hole at (16,3%2F35) is now filled up.

It's graph
is:

graph%28400%2C400%2C-4%2C20%2C-2%2C2%2C%283%28x-16%29%29%2F%28%28x-16%29%282x%2B3%29%29+%29   
 
Edwin