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Now, let's get back to parabolas. There is a basic procedure we can always use to get a pretty good sketch of a parabola. Following it is.
1. Determine the vertex. We'll discuss how to determine this shortly. It's quite simple, but there are several methods for finding it and so will be discussed separately.
2. Find the y-intercept, (0, f (0)) .
3. Solve f ( x ) = 0 to determine the x coordinates of the x-intercepts if they exist.
4. Ensure that you've got at least one point to either side of the vertex. It is to ensure we get a somewhat accurate sketch. If the parabola contains two x-intercepts then already we'll have these points. If it contains 0 or 1 x-intercept we can either just plug in another x value or employ the y-intercept and the axis of symmetry to obtain the second point.
5. Sketch the graph. At this point we've gotten sufficient points to get a quite decent idea of what the parabola will look like.
3x+1=7
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We've been talking regarding zeroes of polynomial and why we require them for a couple of sections now. However, we haven't really talked regarding how to actually determine them f
if I have the equation 3(x+4y)+5(2x-y)an then I change it into (3x+12y)+(10x-5y) what property is it
what is t
Scores on a statewide standardized test are normally distributed with a mean of 12.89 and a standard deviation of 1.95. Certificates are given to students whose scores are in the t
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Write this decimal as a percent. .35
4(x)^2-40x+107 in standard form
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