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: Analyze the parabola's features

The graph is a parabola opening upwards (since the coefficient of \(x^2\) is positive for all options). We can check the y - intercept (when \(x = 0\)) and the discriminant (to see if it has real roots) or complete the square to find the vertex.

First, find the y - intercept by plugging \(x = 0\) into each equation:
- For \(y=3x^{2}-2x + 1\), when \(x = 0\), \(y=1\).
- For \(y=3x^{2}-6x + 3\), when \(x = 0\), \(y = 3\).
- For \(y=3x^{2}-7x + 1\), when \(x = 0\), \(y=1\).
- For \(y=3x^{2}-4x-2\), when \(x = 0\), \(y=-2\).

From the graph, the parabola intersects the y - axis at a positive value, so we can eliminate \(y = 3x^{2}-4x - 2\) (since its y - intercept is - 2).

Step2: Check for real roots (discriminant \(D=b^{2}-4ac\))

A parabola that does not cross the x - axis has a discriminant \(D<0\).
- For \(y=3x^{2}-2x + 1\), \(a = 3\), \(b=-2\), \(c = 1\). \(D=(-2)^{2}-4\times3\times1=4 - 12=-8<0\).
- For \(y=3x^{2}-6x + 3\), \(a = 3\), \(b=-6\), \(c = 3\). \(D=(-6)^{2}-4\times3\times3=36 - 36 = 0\) (this parabola touches the x - axis at one point).
- For \(y=3x^{2}-7x + 1\), \(a = 3\), \(b=-7\), \(c = 1\). \(D=(-7)^{2}-4\times3\times1=49 - 12 = 37>0\) (this parabola crosses the x - axis at two points).

From the graph, the parabola does not cross the x - axis (it is above the x - axis), so we need a parabola with \(D<0\) or \(D = 0\)? Wait, the graph as shown has the vertex above the x - axis and does not intersect the x - axis. Wait, the second option \(y=3x^{2}-6x + 3\) can be factored as \(y = 3(x - 1)^{2}\), which has a vertex at \((1,0)\), so it touches the x - axis. The first option \(y=3x^{2}-2x + 1\) has \(D=-8<0\), so it never touches the x - axis. Wait, maybe I made a mistake in the y - intercept. Wait the graph's y - intercept: let's re - check. The first option: when \(x = 0\), \(y = 1\). The second option: when \(x = 0\), \(y=3\). The graph's y - intercept looks like around 1? Wait, maybe the key is to check the vertex.

The vertex of a parabola \(y=ax^{2}+bx + c\) is at \(x=-\frac{b}{2a}\), \(y = f(-\frac{b}{2a})\).

For \(y=3x^{2}-2x + 1\):
\(x=-\frac{-2}{2\times3}=\frac{1}{3}\), \(y=3\times(\frac{1}{3})^{2}-2\times\frac{1}{3}+1=3\times\frac{1}{9}-\frac{2}{3}+1=\frac{1}{3}-\frac{2}{3}+1=\frac{-1 + 3}{3}=\frac{2}{3}>0\)

For \(y=3x^{2}-6x + 3\):
\(x=-\frac{-6}{2\times3}=1\), \(y=3\times1^{2}-6\times1 + 3=3 - 6 + 3=0\) (vertex at \((1,0)\))

For \(y=3x^{2}-7x + 1\):
\(x=\frac{7}{6}\approx1.17\), \(y=3\times(\frac{7}{6})^{2}-7\times\frac{7}{6}+1=3\times\frac{49}{36}-\frac{49}{6}+1=\frac{49}{12}-\frac{98}{12}+\frac{12}{12}=\frac{49 - 98+12}{12}=\frac{-37}{12}<0\) (so this parabola would cross the x - axis)

For \(y=3x^{2}-4x-2\):
\(x=\frac{4}{6}=\frac{2}{3}\), \(y=3\times(\frac{2}{3})^{2}-4\times\frac{2}{3}-2=3\times\frac{4}{9}-\frac{8}{3}-2=\frac{4}{3}-\frac{8}{3}-\frac{6}{3}=\frac{4 - 8 - 6}{3}=\frac{-10}{3}<0\)

The graph shows a parabola that is entirely above the x - axis (no intersection with x - axis) and has a y - intercept around 1. The first option \(y=3x^{2}-2x + 1\) has \(D=-8<0\) (no real roots) and y - intercept 1, vertex at \((\frac{1}{3},\frac{2}{3})\) which is above the x - axis. The second option has a vertex at \((1,0)\) (touches x - axis). The graph as drawn does not touch the x - axis, so the first option is better? Wait, maybe the original graph's y - intercept is 1. Let's check the options again.

Wait the options are:

1. \(y = 3x^{2}-2x + 1\)

2. \(y=3x^{2}-6x + 3\)

3. \(y=3x^{2}-7x + 1\)

4. \(y=3x^{2}-4x-2\)

The graph is a parabola opening upwards, not intersecting the x - axis (since it's above the x - axis). So we need a quadratic with discriminant less than 0.

For \(y=3x^{2}-2x + 1\), \(D=(-2)^2-4\times3\times1=4 - 12=-8<0\) (no real roots, so the parabola does not cross the x - axis)

For \(y=3x^{2}-6x + 3\), \(D=(-6)^2-4\times3\times3=36 - 36 = 0\) (one real root, touches x - axis)

For \(y=3x^{2}-7x + 1\), \(D=49 - 12 = 37>0\) (two real roots, crosses x - axis)

For \(y=3x^{2}-4x-2\), \(D=16+24 = 40>0\) (two real roots, crosses x - axis)

Since the graph does not cross the x - axis, the equation must be \(y = 3x^{2}-2x + 1\)

Answer:

\(y = 3x^{2}-2x + 1\) (the first option)