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Unformatted text preview: rcept : ( 0,10 ) 2 Next we need to find the xintercepts. This means we’ll need to solve an equation. However,
before we do that we can actually tell whether or not we’ll have any before we even start to solve
the equation.
In this case we have a = 2 which is positive and so we know that the parabola opens up. Also
the vertex is a point below the xaxis. So, we know that the parabola will have at least a few
points below the xaxis and it will open up. Therefore, since once a parabola starts to open up it
will continue to open up eventually we will have to cross the xaxis. In other words, there are xintercepts for this parabola.
To find them we need to solve the following equation. 0 = 2 ( x + 3)  8
2 We solve equations like this back when we were solving quadratic equations so hopefully you
remember how to do them. 2 ( x + 3) = 8
2 ( x + 3) 2 =4 x + 3 = ± 4 = ±2
x = 3 ± 2
The two xintercepts are then, Þ ( 5, 0 ) and x = 1, x = 5 ( 1, 0 ) Now, at this point we’ve got points on eithe...
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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
 MrVinh

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