when graphed, which parabola opens downward?\n$y = -3x^2$\n$y = (x - 3)^2$\n$y = \\frac{1}{3}x^2$\n$y = x^2…
when graphed, which parabola opens downward?\n$y = -3x^2$\n$y = (x - 3)^2$\n$y = \\frac{1}{3}x^2$\n$y = x^2 - 3$
Answer
Explanation:
Step1: Recall parabola direction rule
For a quadratic function in the form $y = ax^2 + bx + c$, if $a < 0$, the parabola opens downward; if $a > 0$, it opens upward.
Step2: Identify $a$ for each option
- For $y=-3x^2$: $a=-3$ (negative)
- For $y=(x-3)^2=x^2-6x+9$: $a=1$ (positive)
- For $y=\frac{1}{3}x^2$: $a=\frac{1}{3}$ (positive)
- For $y=x^2-3$: $a=1$ (positive)
Answer:
A. $y = -3x^2$