Question

PLEASE HELP precal!

Answer 1

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"


`\bf y=a(x-0)^2+0\implies y=ax^2\qquad \begin{cases} x=100\\ y=-50 \end{cases}\implies -50=a100^2 \\\\\\ \cfrac{-50}{100^2}=a\implies -\cfrac{1}{200}=a \\\\\\ thus\qquad \qquad y=-\cfrac{1}{200}x^2\implies \boxed{y=-\cfrac{x^2}{200}}`

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.