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: 3
Step-by-step explanation: y=-2, also y=-2x+4. Therefore, we got a equation: -2=-2x+4. You then minus 4 from both sides of the equation and you will get -6=-2x. You then solve for x by dividing -6 by -2, you get x=3
Supplementary angles.
This is because supplementary angles are two angles that add up to 180 degrees, and these two angles would add up to 180 degrees.
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