Questions: Which represents a quadratic function with no x-intercepts? (A) y=-(x-1)^2 (B) y=-(x-1)^2+3 (C) y=(x+1)^2-3 (D) y=(x+1)^2+3

Solution Steps

A quadratic function has no \(x\)-intercepts if its graph does not cross the \(x\)-axis. This occurs when the quadratic equation \(y = 0\) has no real solutions. For a quadratic function in the form \(y = a(x - h)^2 + k\), this happens when the vertex \((h, k)\) lies entirely above or below the \(x\)-axis, and the parabola does not intersect the \(x\)-axis.

Check each option to see if the vertex lies entirely above or below the \(x\)-axis.

• (A) \(y = -(x - 1)^2\):
  The vertex is at \((1, 0)\). Since the vertex lies on the \(x\)-axis, the parabola touches the \(x\)-axis at one point, so it has one \(x\)-intercept.

• (B) \(y = -(x - 1)^2 + 3\):
  The vertex is at \((1, 3)\). Since the parabola opens downward (due to the negative coefficient) and the vertex is above the \(x\)-axis, the graph does not intersect the \(x\)-axis. Thus, there are no \(x\)-intercepts.

• (C) \(y = (x + 1)^2 - 3\):
  The vertex is at \((-1, -3)\). Since the parabola opens upward and the vertex is below the \(x\)-axis, the graph intersects the \(x\)-axis at two points. Thus, there are two \(x\)-intercepts.

• (D) \(y = (x + 1)^2 + 3\):
  The vertex is at \((-1, 3)\). Since the parabola opens upward and the vertex is above the \(x\)-axis, the graph does not intersect the \(x\)-axis. Thus, there are no \(x\)-intercepts.

From the analysis, options (B) and (D) represent quadratic functions with no \(x\)-intercepts. However, since the question asks for a single correct answer, we select the first valid option, which is (B).

Final Answer