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.
-3 from all ends to get 1<-3x<3 then divide by -3 on all ends to get -1/3>x>-1 you flip the sign because of the negative,
Always add them all around and then divide the two and you should get the correct answer
I'm attaching the solution.. feel free to ask if you have questions.. I basically did long division. Hope this helps.
<em>Answer:</em> ΔTAU ≈ ΔUAV ≈ ΔTUV
<em>Step-by-step explanation:</em>
I'm not really sure what "work" you really need; this is a problem that can be solved easily by simply looking at the triangles and seeing which sides have the same ratio of distances for each side.
Best of luck with your assignment. :) Feel free to give me Brainliest if you feel this helped. Have a good day.
