If a quadratic function has a y-intercept at (0,−1) and an axis of symmetry at x=3 , what is another point on the parabola?

Question

(6, −1)

None of these are correct.

(0, −4)

(−3, −1)

Answer 1

To find another point on the quadratic function given the y-intercept and the axis of symmetry, we can leverage the information provided.

1. Y-Intercept: The point (0, -1) is where the graph crosses the y-axis.

2. Axis of Symmetry: The given axis of symmetry is at \( x = 3 \). This means that the vertex of the parabola is somewhere on the line \( x = 3 \).

Since the y-intercept is to the left of the axis of symmetry and has a value of -1, we can determine that the function is downward-opening. That means there’s a symmetric point on the other side of the axis of symmetry at \( x = 6 \) (which is \(3 + 3\)).

• The y-value at (0, -1) is -1, so we look for another point directly symmetric to this on the right-hand side of \( x = 3 \), which is \( (6, -1) \).

Thus, the point \( (6, -1) \) is indeed on the parabola, and it is symmetric to the point \( (0, -1) \) across the line \( x = 3 \).

So the answer is (6, -1).