008A Sample Final A, Question 10

Question: Graph the function. Give equations of any asymptotes, and list any intercepts $y={\frac {x-1}{2x+2}}$

Foundations:
1) What are the asymptotes and zeros?
Answer:
1) The vertical asymptote corresponds to zeros of the denominator. So there is a vertical asymptote at x = -1. The zero is at (1, 0). The horizontal asymptote is the ratio of leading coefficients. So the horizontal asymptote is $y={\frac {1}{2}}$

Solution:

Step 1:
We start by finding the asymptotes. The vertical asymptote corresponds to zeros of the denominator. So the vertical asymptote is at x = -1. They horizontal asymptote is determined by degree of the numerator and degree of the denominator. Since both of those values are 1, the horizontal asymptote is the ratio of leading coefficients. This means the horizontal asymptote is $y={\frac {1}{2}}$

Step 2:
Now we observe that the zero is at (1, 0), and proceed by looking at the intervals created by removing x = -1 and x = 1. This creates 3 intervals: $(-\infty ,-1)$, $(-1,1)$, and $(1,\infty )$

Step 3:
Now pick a number from each interval: -2, 0, 2 and find the value of the function for each number selected.
$x=-2:$     ${\frac {-2-1}{2(-2)+2}}={\frac {3}{2}}$
$x=0:$     ${\frac {-1}{2}}$
$x=2:$     ${\frac {2-1}{2(2)+2}}={\frac {1}{6}}$

Step 4:
The last check is whether or not the function intersects its horizontal asymptote. So check: ${\frac {1}{2}}={\frac {x-1}{2x+2}}$.
${\begin{array}{rcl}{\frac {1}{2}}&=&{\frac {x-1}{2x+2}}\\2x+2&=&2(x-1)\\2x+2&=&2x-2\\4&=&0\end{array}}$
Since this is absurd, the function never intersects its horizontal asymptote. Now we graph while respecting the asymptotes

Final Answer:

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