To solve this problem, we need to determine which set of points satisfy both the linear equation (unfortunately not visible here but likely implied by the points) and the quadratic equation $y = 4x^2 + 2x - 1$.

Step-by-step solution:

1. Quadratic equation verification: Substitute each $(x, y)$ pair from the options into $y = 4x^2 + 2x - 1$ to check if the point lies on the parabola.

Option A:

• Point 1: $(0, 2)$ $y = 4(0)^2 + 2(0) - 1 = -1 \neq 2$ $(0, 2)$ does NOT lie on the parabola.

• Point 2: $\left(-\frac{1}{2}, 3\right)$ $y = 4\left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right) - 1 = 4\left(\frac{1}{4}\right) - 1 - 1 = 1 - 1 = 0 \neq 3$ $\left(-\frac{1}{2}, 3\right)$ does NOT lie on the parabola.

Option A is eliminated.

Option B:

• Point 1: $\left(\frac{1}{2}, 1\right)$ $y = 4\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) - 1 = 4\left(\frac{1}{4}\right) + 1 - 1 = 1 + 1 - 1 = 1$ $\left(\frac{1}{2}, 1\right)$ lies on the parabola.

• Point 2: $(-1, 4)$ $y = 4(-1)^2 + 2(-1) - 1 = 4(1) - 2 - 1 = 4 - 3 = 1 \neq 4$ $(-1, 4)$ does NOT lie on the parabola.

Option B is eliminated.

Option C:

• Point 1: $\left(-\frac{1}{2}, 3\right)$ $y = 4\left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right) - 1 = 4\left(\frac{1}{4}\right) - 1 - 1 = 1 - 1 = 0 \neq 3$ $\left(-\frac{1}{2}, 3\right)$ does NOT lie on the parabola.

• Point 2: $\left(-\frac{3}{2}, 5\right)$ $y = 4\left(-\frac{3}{2}\right)^2 + 2\left(-\frac{3}{2}\right) - 1 = 4\left(\frac{9}{4}\right) - 3 - 1 = 9 - 3 - 1 = 5$ $\left(-\frac{3}{2}, 5\right)$ lies on the parabola.

Option C is eliminated.

Option D:

• Point 1: $\left(\frac{1}{2}, 1\right)$ $y = 4\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) - 1 = 4\left(\frac{1}{4}\right) + 1 - 1 = 1$ $\left(\frac{1}{2}, 1\right)$ lies on the parabola.

• Point 2: $\left(-\frac{3}{2}, 5\right)$ $y = 4\left(-\frac{3}{2}\right)^2 + 2\left(-\frac{3}{2}\right) - 1 = 4\left(\frac{9}{4}\right) - 3 - 1 = 9 - 3 - 1 = 5$ $\left(-\frac{3}{2}, 5\right)$ lies on the parabola.

Option D is correct.

Final Answer:

D. $\left(\frac{1}{2}, 1\right)$ and $\left(-\frac{3}{2}, 5\right)$

Would you like a detailed explanation of verifying solutions or the derivation process? Here are some related questions to expand your understanding:

1. How do you find solutions to a system involving a linear and a quadratic equation?
2. What methods can be used to solve systems of nonlinear equations graphically?
3. How does substitution simplify solving these systems algebraically?
4. Can a system involving a quadratic and a linear equation have more than two solutions?
5. Why do some points fail to satisfy one of the equations in a system?

Tip: Always substitute each point into both equations when solving a system of equations!