Sketching a parabola from vertex form (a=1)

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CorbettMaths on drawing a parabolaQuestions PDF · Answers
Dr Austin Maths - Plotting Quadratic Graphs Practice GridQuestions PDF · Answers

Read the vertex (h, k) and the axis of symmetry x = h directly from a parabola in vertex form y = (x - h)^2 + k.
State whether the parabola opens up or down and whether it has a minimum or maximum, and find the minimum y-value.
Write the equation of a parabola in vertex form given its vertex, or given its axis of symmetry and minimum value.
Evaluate y for a given x-value, and use symmetry about the vertex to explain why two different x-values give the same y-value.
Find the y-intercept of a vertex-form parabola by substituting x = 0, and find its x-intercepts by solving (x - h)^2 + k = 0.
Determine, from the sign of k, whether a vertex-form parabola has zero, one, or two x-intercepts, and justify the reasoning.
Form a parabola's equation in vertex form from its roots or from its vertex and one other point, and convert between vertex form and factorised form.
Find an unknown parameter (h or k) in a vertex-form equation given a point on the graph or a known root.

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A parabola in vertex (completed-square) form y = (x - h)^2 + k has its vertex directly readable as the point (h, k) -- no completing the square or extra working needed.
The axis of symmetry is the vertical line x = h, passing through the vertex. Since a = 1 is positive, the parabola opens upward and the vertex is a minimum point.
x-intercepts (if any) are found by setting y = 0 and solving (x - h)^2 = -k: real solutions exist only when k <= 0, giving x = h +/- sqrt(-k). If k > 0 the parabola sits entirely above the x-axis and has no x-intercepts.
The y-intercept is found by substituting x = 0, giving y = h^2 + k.