Using the recurrence relation, we can find a couple more values in the sequence:
- a3 = 3a2 -3a1 +a0 = 3(4) -3(2) +2 = 8
- a4 = 3a3 -3a2 +a1 = 3(8) -3(4) +2 = 14
First differences are 0, 2, 4, 6, ...
Second differences are constant at 2, so the function is quadratic.
The sequence can be described by the quadratic ...
... an = n² -n +2
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We know the value for n=0 is 2, so we can find <em>a</em> and <em>b</em> using the given values for a1 and a2.
... an = an² +bn +2
... a1 = 2 = a·1² +b·1 +2 . . . . for n=1
... a + b = 0
... a2 = 4 = a·2² -a·2 +2 . . . . for n = 2; using b=-a from the previous equation
... 2 = 2a
... a = 1 . . . . so b = -1
Answers:
- Vertex form: y = -2(x-1)^2 + 8
- Standard form: y = -2x^2 + 4x + 6
Pick whichever form you prefer.
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Explanation:
The vertex is the highest point in this case, which is located at (1,8).
In general, the vertex is (h,k). So we have h = 1 and k = 8.
One root of this parabola is (-1,0). So we'll plug x = -1 and y = 0 in as well. As an alternative, you can go for (x,y) = (3,0) instead.
Plug those four values mentioned into the equation below. Solve for 'a'.
y = a(x-h)^2 + k
0 = a(-1-1)^2+8
0 = a(-2)^2+8
0 = 4a+8
4a+8 = 0
4a = -8
a = -8/4
a = -2
The vertex form of this parabola is y = -2(x-1)^2+8
Expanding that out gets us the following
y = -2(x-1)^2+8
y = -2(x^2-2x+1)+8
y = -2x^2+4x-2+8
y = -2x^2+4x+6 .... equation in standard form
now, is it true that -2 is lesser than 1? if it's true, is a solution, if is false, is not then
