Question

Which equation could generate the curve in the graph below?

On a coordinate plane, a parabola has a vertex in quadrant 1 and also goes into quadrant 2. The parabola opens up.
y = 3x2 – 2x + 1
y = 3x2 – 6x + 3
y = 3x2 – 7x + 1
y = 3x2 – 4x – 2

Answer 1

We need a parabola that opens up, has its vertex with x>0 and y>0, and also takes positive y for some x<0 (so it enters quadrant II).

Vertex x-coordinate is h = -b/(2a) with a = 3. Check each:

1) y = 3x^2 − 2x + 1: h = −(−2)/6 = 1/3, y(h) = 3(1/9) − 2(1/3) + 1 = 2/3 > 0 → vertex (1/3, 2/3) in Q1. Also f(−1) = 6 > 0 so it goes into Q2. ✓

2) y = 3x^2 − 6x + 3: h = 1, y(h) = 0 → vertex on the x-axis (not Q1).

3) y = 3x^2 − 7x + 1: h = 7/6, y(h) < 0 → vertex below x-axis.

4) y = 3x^2 − 4x − 2: h = 2/3, y(h) < 0 → vertex below x-axis.

So the correct equation is y = 3x^2 − 2x + 1.