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Unformatted text preview: xponent in the numerator and denominator is 1 and so by the fact there will be a
horizontal asymptote at the line. y= 3
1 Now, we just need points each region of x’s. Since the y-intercept and x-intercept are already in
the left region we won’t need to get any points there. That means that we’ll just need to get a
point in the right region. It doesn’t really matter what value of x we pick here we just need to
keep it fairly small so it will fit onto our graph. f ( 2) = 3 ( 2 ) + 6 12
1 ( 2,12 ) Þ Okay, putting all this together gives the following graph. Note that they asymptotes are shown as dotted lines. Example 2 Sketch the graph of the following function.
f ( x) = 2
Okay, we’ll start with the intercepts. The y-intercept is, f ( 0) = 9
-9 Þ ( 0, -1) The numerator is a constant and so there won’t be any x-intercepts since the function can never be
Next, we’ll have vertical asymptotes at, x2 - 9 = 0 Þ x = ±3 So, in this case we’ll have three regions to our graph : x < -...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
- Spring '12