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Question 243395: Trting to assist 13 yr old son...I know very little about algebra.
Find the domain of this rational function.
1)f(x)=5x+6/2x-6
2)f(x)=3x-8/14x+20
Thanks for any assistance
Christine
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! PROBLEM NUMBER 1
f(x)=5x+6/2x-6
The domain of function number 1 is all real values of x except where the denominator is equal to 0.
The denominator is equal to 0 when (2x-6) = 0
Solve for x in 2x-6 = 0
add 6 to both sides to get 2x = 6
divide both sides by 2 to get x = 3
the domain becomaes all real values of x except x = 3.
the range is all real values of y because there does not appear to be any restriction on the values of y.
a graph of this equation looks like this:
The equation has a vertical asymptote at x = 3.
The value of y approaches negative infinity as x approaches 3 from below 3.
the value of y approaches positive infinity as x approaches 3 from above 3.
This particular equation looks like it has a horizontal asymptote at y = 5/2.
the following graph might show that clearer.
the horizontal line at 5/2 shows that the horizontal asymptote appears to be at y = 5/2.
But, you were not asked to show that.
You were only asked to find the domain.
The domain is all real values of x except at x = 3.
PROBLEM NUMBER 2
f(x)=3x-8/14x+20
It looks like the domain here is restricted by when the denominator is equal to 0 again.
The denominator is equal to 0 when 14x + 20 is equal to 0.
14x + 20 is equal to 0 when x = ???
We have to solve for x when 14x + 20 = 0
14x + 20 = 0
subtract 20 from both sides to get 14x = -20
divide both sides by 14 to get x = -20/14 = -10/7
It looks like the domain will be all real values of x except when x = -10/7
The two biggies for restricting the domain are square root of a negative number and division by 0.
This problem deals with division of 0 only.
A graph of this equation looks like this:
It's hard to see, but there is a vertical asymptote at x = -10/7 which is roughly equivalent to -1 and - 3/7, or -1.428571429.
This graph also looks like it has a horizontal asymptote at y = 3/14 but, again, you don't need to worry about that for the questions you are being asked.
If you need to brush up on horizontal or vertical asymptotes, the following website might be helpful.
Horizontal and Vertical Asymptores
Vertical asymptotes are fairly easy.
You just have to find when the denominator is equal to 0.
Horizontal asymptotes are a little more difficult because you have to take the limit of x as x approaches infinity.
In your first equation, what I did was this:
5x+6/2x-6 becomes 5x/2x as x approaches infinity because the 6 and the -6 become insignificant as x gets very large.
5x/2x becomes 5/2 because the x cancels out.
I did a similar procedure for your second equation.
Your domain, however, was a different question.
It was what value of x can be used.
In general, all real values of x can be used except where the equation is not defined for real values of x.
The equation is not defined when the denominator is equal to 0.
The equation is not defined when you wind up with the square root of a negative number.
Those are the two most common causes.
Consider y = square root of x-3
when x is smaller than 3, the square root becomes negative which is not a real number so the domain is restricted to values of x greater than or equal to 3.
a graph of this equation would look like this:
you can see that the graph is not defined for any values of x smaller than 3.
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