Alg_Complete

# So if an intercept crosses the x axis we will call it

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Unformatted text preview: need to take the coordinates of the point and plug them into the equation to see if they satisfy the equation. Let’s do that for both to verify the claims made above. ( 2, -3) : In this case we have x = 2 and y = -2 so plugging in gives, -3 = ( 2 - 1) - 4 ? 2 -3 = (1) - 4 ? 2 -3 = -3 OK So, the coordinates of this point satisfies the equation and so it is a point on the graph. (1, 5) : Here we have x = 1 and y = 5 . Plugging these in gives, 5 = (1 - 1) - 4 ? 2 5 = (0) - 4 ? 2 5 ¹ -4 NOT OK The coordinates of this point do NOT satisfy the equation and so this point isn’t on the graph. Now, how do we sketch the graph of an equation? Of course, the answer to this depends on just how much you know about the equation to start off with. For instance, if you know that the equation is a line or a circle we’ve got simple ways to determine the graph in these cases. There are also many other kinds of equations that we can usually get the graph from the equation without a lot of work. We will see many of these in the next chapter. However, let’s suppose that...
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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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