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Find the equation of the axis of symmetry of the following parabola algebraically. y=4x^(2)+48 x+160 Answer:

Find the equation of the axis of symmetry of the following parabola algebraically.\newliney=4x2+48x+160 y=4 x^{2}+48 x+160 \newlineAnswer:

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Q. Find the equation of the axis of symmetry of the following parabola algebraically.\newliney=4x2+48x+160 y=4 x^{2}+48 x+160 \newlineAnswer:
  1. Identify Quadratic Equation: We have the quadratic equation in the form y=ax2+bx+cy = ax^2 + bx + c, where a=4a = 4, b=48b = 48, and c=160c = 160. To find the axis of symmetry, we use the formula x=−b2ax = -\frac{b}{2a}.
  2. Substitute Values: Substitute the values of aa and bb into the formula: x=−482×4x = -\frac{48}{2 \times 4}.
  3. Calculate: Calculate the value: x=−488x = -\frac{48}{8}.
  4. Simplify Fraction: Simplify the fraction to find the xx-coordinate of the vertex: x=−6x = -6.
  5. Find Axis of Symmetry: The axis of symmetry is a vertical line that passes through the vertex of the parabola. Since the xx-coordinate of the vertex is −6-6, the equation of the axis of symmetry is x=−6x = -6.

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