Check the picture below
so.. .hmmm the vertex is at the origin... and we know the parabola passes through those two points... let's use either.. say hmmm 100,-50, to get the coefficient "a"
keep in mind that, the parabolic dome is vertical, thus we use the y = a(x-h)²+k version for parabolas, which is a vertical parabola
as opposed to x = (y-k)²+h, anyway, let's find "a"

now.. .your choices, show.... a constant on the end.... a constant at the end, is just a vertical shift from the parent equation, the equation we've got above.. is just the parent equation, since we used the origin as the vertex, it has a vertical shift of 0, and thus no constant, but is basically, the same parabola, the one in the choices is just a shifted version, is all.
D is correct.
Inputting any of the numbers under the x column into f(x) =|x| will output the number as a positive
The value of x is 1.
The value of y is 4.
Solution:
Given TQRS is a rhombus.
<u>Property of rhombus:
</u>
Diagonals bisect each other.
In diagonal TR
⇒ 3x + 2 = y + 1
⇒ 3x – y = –1 – – – – (1)
In diagonal QS
⇒ x + 3 = y
⇒ x – y = –3 – – – – (2)
Solve (1) and (2) by subtracting
⇒ 3x – y – (x – y) = –1 – (–3)
⇒ 3x – y – x + y = –1 + 3
⇒ 2x = 2
⇒ x = 1
Substitute x = 1 in equation (2), we get
⇒ 1 – y = –3
⇒ –y = –3 – 1
⇒ –y = –4
⇒ y = 4
The value of x is 1.
The value of y is 4.