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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.
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what is 2+2?
find the perimeter of an irregulary shapep blocks of land didvided into 4 Triangles ab=12m by 15m bc =15m by 60m cd =24m by25m da = 25m by 48m..
2x-y=32 2x+y=60
y=3x+1 x=3y+1
how do you do this im failing mmath
c^-3|d^-8
2/3N + 4 = -26 solve for n, but what do i do with the fraction?
1/2x-2y=4 to slope intercept form
Complete the square on each of the following. x 2 -16x Solution x 2 -16x Here's the number which we'll insert to th
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