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

Answer:

A

Step-by-step explanation:

Answer 2

The correct quadratic equation is the first option: y = 3x^2 - 2x + 1.

Which equation is the one of the given parabola?

We have a parabola with the vertex on quadrant 1, that opens up.

So there are no real roots, meaning that the discriminant must be negative.

Remember that for a quadratic equation:

a*x^2 + b*x + c

The discriminant is:

d = b^2 - 4ac

So let's see the discriminants for the given options.

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

d = (-2)^2 - 4*3*1 = 4 - 12 = -6

2) y = 3x^2 - 6x + 3

d = (-6)^2 - 4*3*3 = 36 - 36 = 0

We can discard this one.

3) y = 3x^2 - 7x + 1

d = (-7)^2 - 4*3*1 = 49 - 12 = 37

We can discard this one.

4) y= 3x^2 - 4x - 2

d = (-4)^2 - 4*3*(-2) = 16 + 24 = 40

We can discard this one.

So, only be checking the discriminant, we can conclude that the only option that can be the correct one is the first option.

y = 3x^2 - 2x + 1

The graph can be seen below, and there you can see that it meets all the conditions.

If you want to learn more about quadratic functions, you can read:

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