Does "b" mean the y-intercept? That's all I can guess it means. The y-intercept is where a graph crosses the y-axis. Points on the y-zxis have x-coordinates of 0. So we substitute 0 in for x to find the y-intercept. We should find that F(0) = -2/-6 = 1/3. So the y-intercept is (0, 1/3).
Your post does not mention a vertical asymptote. But that is an important part of graphing a rational function. Vertical asymptotes, if any, occur for x values that make the denominator zero. TO find them, just set the denominator to zero and solve:
x - 6 = 0
Solving this we should find that x = 6 is a vertical asymptote for F(x).
For the slant asymptote it helps to divide the numerator by the denominator. With a denominator of x-6, this division can be done with synthetic division. (Long division can also be used but I prefer using synthetic division when possible.) Dividing by using synthetic division:
6 | 1 -1 -2
== 6 30
=============
1 5 28
So
Looking at the divided version of F(x), we can tell what will happen when x gets to be very large positive and negative numbers. When x gets to be very large, positive or negative, the denominator of the fraction will become very large (positive or negative). And a fraction with a very large denominator is a very small fraction, close to zero. So for large x's we can ignore the fraction at the end because it will contribute only a very, very small part of the value of F(x). So for large x's, F(x) is essentially going to be x + 5. y = x + 5 is the equation of a line. This line is the slant asymptote for F(x) because for large x's, the value of F(x) will be very close to x + 5. (In fact, the larger x gets the closer F(x) gets to x + 5 since that fraction just keeps getting smaller and smaller as x get bigger.).
From looking at the divided version of F(x) we can tell that for large positive x's, that fraction is a very tiny positive fraction. So F(x) will be close to but a little bit more than the x + 5 of the asymptote for large positive x's (out to the right). And for large negative x's that fraction will be a timy negative fraction. So F(x) will be a little below the asymptote for large negative x's (off to the left).
From the divided version of F(x) we can also tell that F(x) will never cross the slant asymptote. (Some functions will intercept their slant asymptotes.) F(x) will not intercept its slant asymptote since that fraction at the end can never be zero. (It can be extremely close to zero when the denominator is very large. But the only way a fraction can be zero is if its numerator is zero. And the fraction can never be zero since the numerator is a 28 which will never be zero no matter what value x is.)
With this little bit of information we could draw a crude graph. To supplement this information you could:- (If you know some Calculus) Use the first and second derivatives of F(x) to find relative maximums, relative minimums and other aspect of the curvature of F(x).
- Find more points. Just pick some values for x (except 6) and use F(x) to find its y-coordinate. Then plot these points. Do this until you have a sense that you can see how the graph will look. You might want to try some x's near 6 so you can see what happens to F(x) near the vertical asymptote.
To draw the graph:- Draw the two asymptotes, vertical and slant, as dotted lines. (They are dotted because they are not officially part of F(x). They are just guidelines we use to help us draw F(x).
- Plot the points you have found (intercepts, maximums, minimums, others).
- Draw a smooth curve that connects these points. (Keep in mind that we have figured out that
- on the far right of the graph F(x) should be a little above the slant asymptote
- on the far left F(x) should be a little below the slant asymptote
- F(x) should never cross the slant asymptote.
You should get a graph something like:
Notes on the graph:- Algebra.com's graphing software is not perfect. It may look like this graph intersects the asymptotes. The real graph of F(x) will not cross either one.
- I do not know how to make the graphing software draw a dotted line so the asymptotes are not drawn.