PLEASE HELP ME!!! I WILL MARK THE BRAINLIEST!! A given rational function has a quadratic numerator and a linear denominator. When the function is simplified, the denominator cancels out. The graph will look like:
a- a straight line with a hole
b- two branches with a vertical asymptote and a horizontal asymptote
c- two branches with a vertical asymptote, a horizontal asymptote and a hole
d- a straight line with no hole

Answer:

D A strait line with a hole because when you cancel you will leave a hole

Step-by-step explanation:

b

Graph the following:

y = (x^2 - x - 2) / (x - 2)

First I'll find any vertical asymptotes, by setting the denominator equal to zero and solving:

x – 2 = 0

x = 2

So I have a vertical asymptote

at x = 2.

I'll dash this in:

graph showing vertical asymptote

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Looking at the degrees of the numerator and denominator, I see that the numerator is a quadratic while the denominator is linear. Since the degree is one greater in the numerator, I know that I will have a slant asymptote. But I need to do the long division to find out what the equation of the slant asymptote is going to be.

long division: (x^2 - x - 2) ÷ (x - 2) = x + 1

Hmm... The slant asymptote is clearly y = x + 1, but there is no remainder; the division came out evenly. That's because the numerator happens to factor as (x – 2)(x + 1), so the x – 2 factor divides out evenly and there is no remainder.

In any case, the slant (not horizontal) asymptote is at y = x + 1:

graph showing slant asymptote

Now I'll find the intercepts: Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved

x = 0: y = (0 – 0 – 2)/(0 – 2) = 1

y = 0: 0 = (x2 – x – 2) / (x – 2)

0 = x2 – x – 2

0 = (x – 2)(x + 1)

x = 2, x = –1

I know that I can't have x = 2, as an x-intercept, because this is actually the vertical asymptote. So I'll ignore that "solution" of the above equation.

The intercepts then are

(0, 1) and (–1, 0):

graph showing intercepts

Now I'll plot a few more points. However, since this example is actually a special case of slant-asymptote problem (and something of a trick question), I'll also show you the points for the slant asymptote only.

x y = (x2 – x – 2)/(x – 2) y = x + 1

–6 y = (36 + 6 – 2)/(–6 – 2) = –5 y = –5

–5 y = (25 + 5 – 2)/(–5 – 2) = –4 y = –4

–4 y = (16 + 4 – 2)/(–4 – 2) = –3 y = –3

–3 y = (9 + 3 – 2)/(–3 – 2) = –2 y = –2

–2 y = (4 + 2 – 2)/(–2 – 2) = –1 y = –1

–1 y = (1 + 1 – 2)/(–1 – 2) = 0 y = 0

0 y = (0 – 0 – 2)/(0 – 2) = 1 y = 1

1 y = (1 – 1 – 2)/(1 – 2) = 2 y = 2

2 — y is not defined — y = 3

3 y = (9 – 3 – 2)/(3 – 2) = 4 y = 4

4 y = (16 – 4 – 2)/(4 – 2) = 5 y = 5

Do you see how that, except for the one point where the rational function isn't defined (at x = 2), the two lines are the same? In general, this is not true for rationals. But in this special case — where the long division works out with no remainder — you don't actually even need the asyptote lines dashed in. Instead, you find the slant asymptote equation (in this case, y = x + 1), and you draw that in for the rational graph.

The only difference between the slant asymptote of the rational function and the rational function itself is that the rational function isn't defined at x = 2. To account for this, I leave a nice big open circle at the point where x = 2, showing that I know that this point is not actually included on the graph, because of the zero in the denominator of the rational.

graph of y = (x^2 - x - 2) / (x - 2)

Warning: This graph is not the norm for rationals. But you should expect to encounter one like this, perhaps on the test. I've only ever seen this done with slant asymptotes, but it could be done with horizontal asymptotes, too. Keep in mind that you can't just "cancel off" duplicate factors when graphing rationals; you still need to account for all the zeroes of the denominator.

Note that your calculator will probably not show this hole in the graph, and "TRACE" will probably miss it too, since it is unlikely that x = 2 will be exactly on the pixel that the calculator is processing. But even if the graph does not show the gap in the line, the calculator will show no value for y when x is two.

When I asked my calculator to evaluate y for x = 2, I got this:

screen shot showing apparently connected line

If it just so happens that the vertical asymptote falls on a pixel (rather than between two pixels), then the gap in the graph will display on the calculator's pretty picture.

For instance, when I changed the window settings on my graphing calculator so the x-values ran from –7 to 7 (rather than the default –10 to 10), the gap was (just barely) visible:

screen shot showing gap in graph

When you do these graphs, get in the habit of working methodically through all the steps (vertical asymptote(s), horizontal or slant asymptote, intercept(s), plotted points, and graph), so you can do the graphing easily when you get to the test. And, as you can see above, you will need to know this material well enough that your calculator doesn't mess you up.

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