Question

which equation could generate the curve in the graph below?
options:

• $y = 3x^2 - 2x + 1$
• $y = 3x^2 - 6x + 3$
• $y = 3x^2 - 7x + 1$
• $y = 3x^2 - 4x - 2$

Explanation:

Step1: Identify parabola properties

The parabola opens upward (so coefficient of $x^2$ is positive, true for all options) and has a positive y-intercept (when $x=0$, $y>0$).

Step2: Check y-intercepts

• For $y=3x^2-2x+1$: $x=0$, $y=1>0$
• For $y=3x^2-6x+3$: $x=0$, $y=3>0$
• For $y=3x^2-7x+1$: $x=0$, $y=1>0$
• For $y=3x^2-4x-2$: $x=0$, $y=-2<0$ (eliminate this option)

Step3: Find vertex x-coordinate

Vertex x-coordinate is $x=-\frac{b}{2a}$. For $a=3$, $x=-\frac{b}{6}$. The vertex is to the right of the y-axis, so $x>0$, meaning $b<0$ (all remaining options satisfy this). Now check if vertex y-coordinate is positive (graph does not cross x-axis, so discriminant $b^2-4ac<0$):

• For $y=3x^2-2x+1$: $\Delta=(-2)^2-4(3)(1)=4-12=-8<0$ (no real roots, matches graph)
• For $y=3x^2-6x+3$: $\Delta=(-6)^2-4(3)(3)=36-36=0$ (one real root, parabola touches x-axis, does not match graph)
• For $y=3x^2-7x+1$: $\Delta=(-7)^2-4(3)(1)=49-12=37>0$ (two real roots, crosses x-axis, does not match graph)

Answer:

$\boldsymbol{y=3x^2-2x+1}$