Q. Which graph represents the function \(f(x) = (x-5)^2 + 3\)?

Recognize the parent function and the transformations.

The given function is \(f(x) = (x-5)^2 + 3\). The parent function is \(g(x)=x^2\). The form \( (x-h)^2 + k \) indicates a horizontal shift by \(h\) and a vertical shift by \(k\). Here \(h=5\) and \(k=3\).

Conclusion: start with the parabola \(y=x^2\), shift it 5 units to the right, then shift it 3 units up.

• Find the vertex.For a parabola in vertex form \( (x-h)^2 + k \), the vertex is exactly \((h,k)\). Therefore the vertex is \((5,3)\).

  When identifying the correct graph, look for a parabola whose lowest point (since it opens upward) is at \((5,3)\).

• Determine the direction the parabola opens.The coefficient of \((x-5)^2\) is positive (1), so the parabola opens upward. The correct graph must be a U-shaped curve opening upward, with its minimum at the vertex \((5,3)\).
• Find the axis of symmetry.The axis of symmetry is the vertical line through the vertex: \(x=5\). The graph should be symmetric about the line \(x=5\).
• Check intercepts and sample points to distinguish similar graphs.Compute the y-intercept by evaluating at \(x=0\):

  \(f(0) = (0-5)^2 + 3 = 25 + 3 = 28\).

  So the graph must pass through \((0,28)\). This is a very large positive y-intercept, which helps eliminate graphs whose y-intercept is much lower.

  Compute a couple of symmetric points near the vertex: let \(x=4\) and \(x=6\).

  \(f(4) = (4-5)^2 + 3 = 1 + 3 = 4\). Therefore \((4,4)\) is on the graph.

  \(f(6) = (6-5)^2 + 3 = 1 + 3 = 4\). Therefore \((6,4)\) is on the graph.

  These points confirm the parabola is narrow in the vicinity of the vertex and symmetric about \(x=5\).

• Check for real x-intercepts.Solve \( (x-5)^2 + 3 = 0\). This gives \( (x-5)^2 = -3\), which has no real solutions. Therefore there are no x-intercepts; the parabola does not cross the x-axis.

  When picking the graph, eliminate any candidate that crosses the x-axis.

• Summary of identifying features (use these to choose the correct graph):
  • Vertex at \((5,3)\).
  • Parabola opens upward.
  • Axis of symmetry \(x=5\).
  • Y-intercept at \((0,28)\).
  • Points \((4,4)\) and \((6,4)\) lie on the curve.
  • No real x-intercepts (does not cross the x-axis).
  • Minimum value is \(3\) at \(x=5\).
• How to use this to select the correct graph:From the candidate graphs, choose the one that matches all the features above: a U-shaped curve opening upward with its lowest point at \((5,3)\), symmetric about \(x=5\), passing through \((4,4)\) and \((6,4)\), having a very high y-intercept at \((0,28)\), and not intersecting the x-axis.