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if the parabola of the form $y = a(x - h)^2 + k$ is always shifted hori…

Question

if the parabola of the form $y = a(x - h)^2 + k$ is always shifted horizontally $h$ units and vertically $k$ units, then its vertex is always\
\\(\circ\\) $(-h, -k)$\
\\(\circ\\) $(h, k)$\
\\(\circ\\) $(-h, k)$\
\\(\circ\\) $(h, -k)$\
done

Explanation:

Step1: Recall Vertex Form of Parabola

The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \((h, k)\) represents the vertex of the parabola. This is because the standard form \( y = ax^2 \) has its vertex at \((0, 0)\). When we replace \( x \) with \( x - h \), the graph shifts horizontally by \( h \) units (right if \( h>0 \), left if \( h<0 \)), and adding \( k \) shifts the graph vertically by \( k \) units (up if \( k>0 \), down if \( k<0 \)). So the vertex, which was at \((0, 0)\) for \( y = ax^2 \), moves to \((h, k)\) after these shifts.

Step2: Analyze the Options

  • Option \((-h, -k)\): Incorrect, as the shifts are \( h \) (horizontal) and \( k \) (vertical), not their negatives.
  • Option \((h, k)\): Correct, as per the vertex form definition.
  • Option \((-h, k)\): Incorrect, horizontal shift is \( h \), not \(-h\).
  • Option \((h, -k)\): Incorrect, vertical shift is \( k \), not \(-k\).

Answer:

B. \((h, k)\) (assuming the second option is labeled B, with text \((h, k)\))