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y=-4(x-2)^(2)-1 Vertex: Axis of symmetry: Range:

77. y=−4(x−2)2−1 y=-4(x-2)^{2}-1 \newlineVertex:\newlineAxis of symmetry:\newlineRange:

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Q. 77. y=−4(x−2)2−1 y=-4(x-2)^{2}-1 \newlineVertex:\newlineAxis of symmetry:\newlineRange:
  1. Vertex Form Explanation: The vertex form of a parabola is y=a(x−h)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex.\newlineFor y=−4(x−2)2−1y = -4(x - 2)^2 - 1, the vertex is (h,k)=(2,−1)(h, k) = (2, -1).
  2. Axis of Symmetry: The axis of symmetry is the line that passes through the vertex and is parallel to the y-axis.\newlineSo, the axis of symmetry is x=hx = h, which is x=2x = 2.
  3. Range Determination: The range of the function depends on the direction the parabola opens and the vertex.\newlineSince the coefficient of (x−2)2(x - 2)^2 is −4-4, the parabola opens downwards.
  4. Downward-Opening Parabola: For a downward-opening parabola, the range is all yy-values less than or equal to the yy-coordinate of the vertex.\newlineSo, the range is y≤−1y \leq -1.

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