Question

When graphed, which parabola opens downward? y = –3x2     y = (x – 3)2    y = x2 – 3   

Answer 1

we know that

the equation of a vertical parabola into vertex form is equal to

$y=a(x-h)^{2} +k$

where

(h,k)-------> is the vertex of the parabola

if $a > 0$ -------> the parabola open upwards

if $a < 0$ -------> the parabola open downwards

case A) $y=-3x^{2}$

$a =-3$

so

$a < 0$ -------> the parabola open downwards

the vertex is the point $(0,0)$ ------> is a maximum

therefore

$y=-3x^{2}$ open downwards

case B) $y=(x-3)^{2}$

$a =1$

so

$a > 0$ -------> the parabola open upwards

the vertex is the point $(3,0)$ --------> is a minimum

therefore

$y=(x-3)^{2}$ open upwards

case C) $y=x^{2}-3$

$a =1$

so

$a > 0$ -------> the parabola open upwards

the vertex is the point $(0,-3)$ --------> is a minimum

therefore

$y=x^{2}-3$ open upwards

therefore

the answer is

$y=-3x^{2}$ open downwards

Answer 2

The answer to this question is y = –3x^2, just had this on e2020 :)