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?
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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?
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Question 176275: My math question reads:
Find the hole in this function
Can you please help me figure this out? Found 2 solutions by stanbon, Edwin McCravy:Answer by stanbon(75887) (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.
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.
Factor the numerator and denominator:
DO NOT cancel the '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 .
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.
Now if you cancel the 's, you
will get a NEW function:
which is exactly like your given function except
the hole will be plugged up. That's because with
the '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:
So the hole in the previous graph is the point (16,).
The graph of the new function
is exactly like your original function except you will notice
that the hole at (16,) is now filled up.
It's graph
is:
Edwin
(