2/x-3+1/x=x-1/x-3
To do so, we will first determine the least common denominator (LCD) so that we can clear the denominators. Note that the denominators are already factored. Therefore, in this case the LCD is the product between (x-3) and x.
Let's calculate both solutions by using the positive and negative signs.
x=4± 2/2
x_1=4+ 2/2
x_2=4- 2/2
x_1=6/2
x_2=2/2
x_1=3
x_2=1
Finally, we will check whether either x=3 or x=1 are extraneous solutions. To do so, we will substitute 3 and 1 for x in the original equation. Let's start with the first one.
Therefore, the function has vertical asymptotes at x=0 and x=3. Next, we will find the horizontal asymptote of the function. To do so, let's pay close attention to the degrees of the numerator and denominator.
y=3x^1-3/x^2-3x
We see that the degree of the denominator is higher than the degree of the numerator. Therefore, the line y=0 is a horizontal asymptote. Now, we will plot the asymptotes.
Next, we will make a table of values. Recall that we are asked to graph the function for x-values greater than 0 andless than 5. Since the function has a vertical asymptote at x=3, this value will be excluded from the domain.
x
3x-3/x(x-3)
y
0.5
3( 0.5)-3/0.5( 0.5-3)
1.2
1
3( 1)-3/1( 1-3)
0
1.5
3( 1.5)-3/1.5( 1.5-3)
≈- 0.67
2
3( 2)-3/2( 2-3)
- 1.5
2.5
3( 2.5)-3/2.5( 2.5-3)
-3.6
3.5
3( 3.5)-3/3.5( 3.5-3)
≈ 4.29
4
3( 4)-3/4( 4-3)
≈ 2.25
4.5
3( 4.5)-3/4.5( 4.5-3)
≈ 1.56
Finally, let's plot and connect the points.
We can graph y= x-1x-3 proceeding in the same way. The function has also a vertical asymptote at x=3. Let's pay close attention to the degrees of the numerator and denominator.
y=x^1-1/x^1-3
We see that the degrees of the numerator and denominator are the same. To find the horizontal asymptote, we need to find the quotient between the leading coefficients.
y=1x-1/1x-3
Since 11=1, there is a horizontal asymptote at y=1. By finding, plotting, and connecting points, we can graph the function as we did for the first one.
c
Let's examine the graph that we drew in Part B.
We see that the graph of the functions intersect at x=1. For the extraneous solution of x=3, the graphs do not intersect. In fact, x=3 is a vertical asymptote.
d
We first graph the functions related to both sides of the equation. If the graphs intersect, then we can say that the point of intersection is a solution. If the graphs do not intersect, then the possible solution is extraneous.
