Answer:
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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.
The answer is B.)
The answers are using the distributive property. The distributive property multiplies both the numbers inside the parenthesis by the number that's on the outside of the parenthesis.
6x3 = 18, 6x4x=24x.
Hope this helps :) work hard!
Answer:
64k^2 - 16k +1
Step-by-step explanation:
We can rewrite this as
(-8k+1) ^2
We know that (a+b)^2 = a^2 +2ab +b^2
Let a = -8k and b = 1
(-8k+1) = (-8k)^2 +2*(-8k)(1) + 1^2
=64k^2 - 16k +1
