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