Sorry but im terrible at math
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:
Step-by-step explanation:
Here are some steps to help you
Step 1
4x-3≥13 We are going to be simplifying
Step 2
4x-3≥13 Add 3 to the sides
4x≥16
Step 3
4x≥16 Divide them sides by 4
x≥4
So therefore your answer is x≥4
Hope this helps
Answer:
12
Step-by-step explanation:
4*x-9=2x+15
4x-9=2x+15
-2x -2x
2x-9=15
2x=24
x=12
