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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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We've some rather simply tests for each of the distinct types of symmetry. 1. A graph will have symmetry around the x-axis if we get an equal equation while all the y's are repl
please help me understand polynomials- i get the small problems but i dont understand larger ones
3 with exponent of x-2=7
Alex lives .4 miles from the park, beth lives .8 miles from the park. To run equal distances Alex runs 8 times around the park and Beth run 6 times around the part. Write an equati
-5t=-45
3x+1=7
2.3*10^3,3.7*10^2,6.5*10^3
find the range of f(x)=2x+4 for the domain {-4,-1,3,4].
Prove that cosets of an ideal I in a ring R are disjoint or equal
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