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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.
if im working out a problem that say 16 - t over 10 = -8 what be the answer
10n+2+32 what does n equal
(1, 5) and (2, 6)
There are also two lines on each of the graph. These lines are called asymptotes and as the graphs illustrates as we make x large (in both the +ve and -ve sense) the graph of the h
52% of 0.40 of =
How do you do uniform motion? i am stuck on some problems
(2x^2+16x+24)/(3x^2) /(x+6)
What do you have to do to be able to answer any problem solving questions (relating to or concerning algebra)?
Next we desire to take a look at f (x ) =√x . First, note that as we don't desire to get complex numbers out of a function evaluation we ought to limit the values of x that we can
Need to solve for x x over 17/5 equals 1/17
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