How do I sketch the graph y= (2x+5)/(x-1)

In summary, to sketch the graph of y= (2x+5)/(x-1), you can first simplify it by multiplying the first constant by the highest common denominator. This will give you the simplified expression of (4x+5)/(2x+1). To go back to the original expression, you need to convert it into the form of (a(2x+1)+b)/(2x+1) by canceling out the x in the numerator. By setting a=2, you can have 4x in the numerator. To keep the numerator the same, you need to find a number b that makes the equation 2(2x+1)+b=4x+5 true. This gives
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How do I sketch the graph y= (2x+5)/(x-1)
 
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Say you want to simplify
[tex]2+\frac{3}{2x+1}[/tex]
then you would multiply the first constant by the highest common denominator so you can put it all in one fraction as so:
[tex]\frac{2(2x+1)+3}{2x+1}[/tex]
and then simplify and you get
[tex]\frac{4x+5}{2x+1}[/tex].
But what if you started with this expression and wanted to go back to the start now?

You need to cancel out the x in the numerator so you need to convert it into the form
[tex]\frac{a(2x+1)+b}{2x+1}[/tex]
Notice that the numerator has 4x in it so a=2 so that we can have this 4x.
[tex]\frac{2(2x+1)+b}{2x+1}[/tex]
Now in order to keep the numerator the same we need a number b so that it makes [itex]2(2x+1)+b=4x+5[/itex] and once you solve this you get b=3.
Then we get
[tex]\frac{2(2x+1)+3}{2x+1}=\frac{2(2x+1)}{2x+1}+\frac{3}{2x+1}=2+\frac{3}{2x+1}[/tex]

From this expression we can easily sketch the graph
[tex]y=2+\frac{3}{2x+1}[/tex]
by noting that there is a vertical asymptote at 2x+1=0, x=-1/2 and there is a horizontal asymptote at y=2 (since the fraction never equals 0).

Now see if you can apply the same idea to your problem.
 

1. How do I determine the domain and range of this graph?

The domain of a function is the set of all possible input values, while the range is the set of all possible output values. To determine the domain, we need to find all values of x that make the denominator (x-1) equal to 0. In this case, x cannot equal 1, so the domain is all real numbers except x=1. To find the range, we can set up a table and plug in different values for x to see what values y can take. Alternatively, we can use the fact that the graph is a hyperbola, and its range is all real numbers except y=2 (since y=2 is the horizontal asymptote).

2. How do I find the x-intercept and y-intercept of this graph?

The x-intercept is the point where the graph crosses the x-axis, which means that y=0. To find the x-intercept, we can set y=0 and solve for x. In this case, we get x=-5/2. The y-intercept is the point where the graph crosses the y-axis, which means that x=0. To find the y-intercept, we can set x=0 and solve for y. In this case, we get y=5.

3. How do I determine the slope and y-intercept of the graph?

The slope of a line is a measure of its steepness, and it is given by the formula m=(y2-y1)/(x2-x1), where (x1,y1) and (x2,y2) are any two points on the line. In the case of this graph, we can rewrite the equation as y=2+7/(x-1), which is in slope-intercept form y=mx+b, where m=2 and b=7. Therefore, the slope is 2 and the y-intercept is (0,7).

4. How do I determine the symmetry of this graph?

A graph can have symmetry with respect to the x-axis, y-axis, or origin. To determine the symmetry of this graph, we can replace x with -x and see if the resulting equation is equivalent to the original equation. If it is, then the graph is symmetric with respect to the y-axis. Similarly, replacing y with -y and seeing if the resulting equation is equivalent to the original equation can determine symmetry with respect to the x-axis. In this case, neither of these replacements yield equivalent equations, so the graph is not symmetric with respect to the x-axis or y-axis. However, replacing x with -x and y with -y yields an equivalent equation, which means that the graph is symmetric with respect to the origin.

5. How do I sketch the graph of this equation?

To sketch the graph of this equation, we can first plot the x and y-intercepts and then use the slope to determine the direction of the graph. In this case, the x-intercept is -5/2 and the y-intercept is 5. Since the slope is positive, the graph will be increasing on both sides of the asymptote x=1. To plot more points, we can choose values for x that are close to 1 and plug them into the equation. We can also use the symmetry property to plot points on the other side of the origin. Once we have enough points, we can connect them to form the graph of the hyperbola.

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