6. Draw the graph of

• a) What is the y-intercept?
• b) What are the roots?
• c) Where are the vertical asymptotes?
• d) Are there any horizontal asymptotes, and if so where are they?
• e) Solve for   x

It will be helpful to answer questions a), b), c), and d) before attempting to draw the graph.

a) To find the y-intercept, let   x = 0.   The equation becomes

= 1
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b) To find the roots or x-intercepts, let   y = 0.   The equation becomes

This gives us a rational equation to solve for x. Clear denominators

(x - 1)0 = 2x - 1

But 0 times anything is 0, so we get

0 = 2x - 1

This illustrates that the only way that a fraction can be 0 is if the top is 0. So the roots are the roots of the top.

1 = 2x

Divide by 2.

1/2 = x

The only x-intercept is when   x = 1/2

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c) Vertical asymptotes. Vertical asymptotes are caused by 0 denominators. Set the denominator = 0.

x - 1 = 0

x = 1

is a vertical asymptote.

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d) In this case the bottom has the same degree as the top. We divide the bottom into the top using long division of polynomials

and we see that

Since the fractional term has a top which has smaller degree than the bottom, it will become negligible,

y = 2

will be a horizontal asymptote.

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When we make up the graph, we start by putting in these intercepts and asymptotes.

When we plot points, we should plot at least one x in each interval between the roots and vertical asymptotes.

The first interval is the set of x's which are less than ½. We know that the y-intercept is (0, 1). This would be a good point in this first interval. The next inteerval is between ½ and 1. The simplest point in this interval is ¾, and we find that the function has a value of -2 there. We know that there is a vertical asymptote when x = 1. We indicate that by having a function value of   ∞   there. The last interval is the set of x's which are greater than 1. The simplest number to use in that interval would be 2, and we find that ther is a function value of 3 there.

When we plot these points in conjunction with our roots and asymptotes, we have to get something that looks something like

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e) Solve for   x

Since we have a 0 on one side and a single fraction on the other, we are ready to find the roots of the top and the bottom.

Roots of the top: Set

2x - 1 = 0

Solve for   x

2x = 1

x = 1/2

The roots of the top are the roots of the function that we got when we drew the graph.

Roots of the bottom: Set

x - 1 = 0

Solve for   x

x = 1

The roots of the bottom are where the vertical asymptotes cross the x-axis. These are the two types of places where the graph of the function can cross the x-axis.

Divide the real number line up into the intervals between the roots of the top and bottom. The intervals will be

(-∞, ½),   (½,1),   and   (1, ∞)

When we pick a point in each interval, we may as well use the points that we used when we drew the graph.

When we check the endpoints of the intervals, since equality is permissible, the roots of the top will be solutions, because those are the x's that give us answers that are equal to zero, but the roots of the bottom are not solutions, because a root of the bottom will give us a denominator of zero when we substitute it in, and if your denominator is 0, then your fraction is not a real number: certainly not 0. So the root of the top, ½ is a solution, and the root of the bottom, 1, is not, so we draw the graph accordingly.

The solution set is the set of all x's that satisfy

x ≤ ½   or   x > 1

The interval notation would be

(-∞, ½] ∪ (1, ∞)

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