Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.
y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).
Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,
-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.
The equation of parabola y4 is y+4 = (1/4)x^2
Or you could elim. the fraction and write the eqn as 4y+16=x^2, or
4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.
Answer: Choice A
S9 = (9/2)*(2+26)
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The formula is
Sn = (n/2)*(a1+an)
where
Sn = sum of the first n terms (nth partial sum)
n = number of terms
a1 = first term
an = nth term
In this case,
n = 9
a1 = 2 (plug in n = 1 into the formula an = 3n-1 and simplify)
an = a9 = 26 (plug n = 9 into the formula an = 3n-1 and simplify)
So,
Sn = (n/2)*(a1+an)
S9 = (9/2)*(2+26)
will help us find the sum of the first 9 terms of this arithmetic sequence
You find the slope of the line by using the formula 
where x1 and y1 is the first point, and x2 and y2 are the second point.
Step-by-step explanation:
Are
Bsushdid djjdu
Dhhduehehid
Dhdhuduhdje
