2007 paul dawkins 159

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Unformatted text preview: for the x-intercept(s) is almost identical except in this case we set y = 0 and solve for x. Here is that work. 0 = x2 + x - 6 0 = ( x + 3)( x - 2 ) Þ x = -3, x = 2 For this equation there are two x-intercepts : ( -3, 0 ) and ( 2, 0 ) . Oh, and you do remember how to solve quadratic equations right? For verification purposes here is sketch of the graph for this equation. [Return to Problems] (b) y = x 2 + 2 First, the y-intercepts. y = ( 0) + 2 = 2 2 Þ ( 0, 2 ) So, we’ve got a single y-intercepts. Now, the x-intercept(s). 0 = x2 + 2 -2 = x 2 Þ x = ± 2i Okay, we got complex solutions from this equation. What this means is that we will not have any x-intercepts. Note that it is perfectly acceptable for this to happen so don’t worry about it when it does happen. Here is the graph for this equation. © 2007 Paul Dawkins 157 http://tutorial.math.lamar.edu/terms.aspx College Algebra Sure enough, it doesn’t cross the x-axis. [Return to Problems] (c) y = ( x + 1) 2 Here is the y-intercept work for this equation. y = ( 0 + 1) = 1 2 ( 0,1) Þ Now the x-intercept w...
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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.

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