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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.
Can you please expalin to me how to solve this question? I am totally lost. A client comes to you for investment advice on his $500,000 winnings from the lottery. He has been off
six is at least four more than a number. write the inequality represents in this sentence
Question 1 [7 marks] The reduced row echelon form of ? ? ? ? ? ? ? ? ? ? ? ? 3 4 17 22 1 2 5 row a b A is equal to ? ? ? ? ? ? ? ? ? ? ? ? 0 0 0 0 0 0 1 2 1 2 0 3 R . (a) What can
Graphing and Functions Graphing In this section we have to review some of the fundamental ideas in graphing. It is supposed that you've seen some graphing at th
x2 -x = 7 -3 -5 1 5 -3
how to find hcf easily
what is the answer for this 2n+3n+7=-41
x2=4
(-9) (-88) (-7)
y-5=2x
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