Question 344488
X intercepts are the point(s), if any, where a graph intersects the x axis. To find them just think about points on the x axis. Think about the coordinates of points on the x axis. It shouldn't take long to realize that <i>all</i> points on the x axis have a y coordinate that is zero! And this exactly how you find x intercepts: Make the y (or function value) zero and solve for x.<br>
Y intercepts are the point(s), if any, where a graph intersects the y axis. To find them just think about points on the y axis. Think about the coordinates of points on the y axis. It shouldn't take long to realize that <i>all</i> points on the y axis have a x coordinate that is zero! And this exactly how you find y intercepts: Make the x zero and solve for y (or function value).<br>
{{{y= (-4)/(x-1)}}}
You have the correct asymptotes. For the x intercept(s), using the logic from above:
{{{0= (-4)/(x-1)}}}
Now we try to solve for x. But there is no solution to this equation. This equation says that a fraction is zero. But a fraction can only be zero if the numerator is zero. The numerator is -4 and can never be zero so this equation has no solutions. This means that {{{y= (-4)/(x-1)}}} has no x intercepts.<br>
For the y intercept(s), using the logic from above:
{{{y= (-4)/(0-1) = (-4)/(-1) = 4}}}
So the only y intercept is (0,4).<br>
At this point we have two asymptotes and a single point. To this we can add some analysis of how the horizontal asymptote will work. For very large positive values of x the y will be -4 over a very large positive x. This gives us a <i>negative</i> number that is very close to zero. So for large positive x's, the graph will approach zero <i>from below</i> (where the negative y's are). For very large negative x's we get -4 over a very large negative number which is a positive number which is very close to zero. So for large negative x's, the graph will approach zero from above (where the positive y's are).<br>
The two asymptotes, the point and the analysis is still not enough to draw a fairly reasonable graph. So we need more points. For these points we pick x values and find their corresponding y values. While you are free to pick any numbers you want for x (except for the x's of any vertical asymptotes), I suggest that you start from the vertical asymptote(s) and work outward.<br>
Your vertical asymptote is x=1. Start from 1 and move outward (to the right and to the left). One to the right would be x=2 and one to the left would be x=0. So find the y values for these x's. (You already have the y for x=0 because this is the y intercept.) For x = 2, y = -4/(2-1) = -4/1 = -4.
Plot these points: (0,4) and (2, -4).<br>
Next try one more to the right and one more to the left: x=3 and x=-1:
y = -4(3-1) = -4/2 = -2
and
y = -4(-1-1) = -4/-2 = 2
Now we have the points (3, -2) and (-1, 2). Plot them.<br>
Now repeat this until you think you have enough points to "see" how the graph goes. (Keep in mind the analysis of the horizontal asymptote.) Probably one or two more pairs will be enough. Then draw a smooth curve which goes through the points you've plotted and which approaches the asymptotes. Your final graph should resemble a slanted hyperbola (because it is one).<br>
For the second problem, {{{y = ((x+2)(x-2))/(x-2)}}} you have correctly determined that it has the graph of y = x+2 with a hole at 2. To draw this graph we need to be able to graph y = x+2 with a hole in it. I recommend that you plot the "hole" first. For a hole at 2, find the y value for x=2 in the "reduced" equation of y = x+2: y = 2+2 = 4. So the "hole" is at the point (2, 4). To graph a "hole" we use an open circle, not a dot like we use for points we want to include. This is <i>exactly</i> like the open circles you use when graphing the solution to x > 4 on a number like!<br>
At this stage, we just have an open circle at (2, 4). Now we just draw the graph of y = x+2 <i>without drawing through the hole!</i> As I hope you recognize, y = x+2 is a line with a slope of 1 and a y-intercept of 2. This should be easy to graph. Just avoid drawing through the hole at (2, 4)! The line should go right up to the hole on each side, giving you a line with a "hole" or gap at (2, 4).