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 (positive leading coefficient, all options have 3, so this is satisfied) and has its vertex in the first quadrant (positive x and positive y for the vertex), and the y-intercept is positive.

Step2: Find vertex x-coordinate

For a quadratic $y=ax^2+bx+c$, vertex x is $x=-\frac{b}{2a}$.

• For $y=3x^2-2x+1$: $x=-\frac{-2}{2*3}=\frac{1}{3}$
• For $y=3x^2-6x+3$: $x=-\frac{-6}{2*3}=1$
• For $y=3x^2-7x+1$: $x=-\frac{-7}{2*3}=\frac{7}{6}$
• For $y=3x^2-4x-2$: $x=-\frac{-4}{2*3}=\frac{2}{3}$

Step3: Check vertex y-coordinate

Calculate y at vertex x, ensure positive, and check y-intercept (c value, must be positive):

• $y=3x^2-2x+1$: $y=3(\frac{1}{3})^2-2(\frac{1}{3})+1=\frac{1}{3}-\frac{2}{3}+1=\frac{2}{3}$, y-intercept $1>0$
• $y=3x^2-6x+3$: $y=3(1)^2-6(1)+3=0$, vertex on x-axis (does not match graph)
• $y=3x^2-7x+1$: $y=3(\frac{7}{6})^2-7(\frac{7}{6})+1=\frac{49}{12}-\frac{49}{6}+1=-\frac{37}{12}$, negative vertex y (does not match)
• $y=3x^2-4x-2$: y-intercept $-2<0$ (does not match graph)

Step4: Verify graph match

The parabola $y=3x^2-2x+1$ has a positive y-intercept, vertex in the first quadrant, and opens upward, matching the graph.

Answer:

$y=3x^2-2x+1$