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Unformatted text preview: Paul Dawkins 176 http://tutorial.math.lamar.edu/terms.aspx College Algebra and see what we get. x = 3: y2 = 3 +1 = 4 Þ y = ±2 x = -1: y 2 = -1 + 1 = 0 Þ y=0 x = 10 : y 2 = 10 + 1 = 11 Þ y = ± 11 Now, remember that we’re solving for y and so that means that in the first and last case above we
will actually get two different y values out of the x and so this equation is NOT a function.
Note that we can have values of x that will yield a single y as we’ve seen above, but that doesn’t
matter. If even one value of x yields more than one value of y upon solving the equation will not
be a function.
What this really means is that we didn’t need to go any farther than the first evaluation, since that
gave multiple values of y.
[Return to Problems] (d) x 2 + y 2 = 4
With this case we’ll use the lesson learned in the previous part and see if we can find a value of x
that will give more than one value of y upon solving. Because we’ve got a y2 in the problem this
shouldn’t be too hard to do since solving will eventually mean using the square root property
which will give more than one value of y. x = 0:...
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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