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Find the coordinates of the vertex of the following parabola using graphing technology. Write your answer as an (x,y) point. y=4x^(2)-16 Answer:

Find the coordinates of the vertex of the following parabola using graphing technology. Write your answer as an (x,y) (x, y) point.\newliney=4x2−16 y=4 x^{2}-16 \newlineAnswer:

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Q. Find the coordinates of the vertex of the following parabola using graphing technology. Write your answer as an (x,y) (x, y) point.\newliney=4x2−16 y=4 x^{2}-16 \newlineAnswer:
  1. Identify Vertex Form: 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 of the parabola. To find the vertex of the parabola given by y=4x2−16y = 4x^2 - 16, we need to complete the square to convert the equation into vertex form.
  2. Factor Out Coefficient: The given equation is y=4x2−16y = 4x^2 - 16. To complete the square, we need to factor out the coefficient of x2x^2 from the xx-terms.
  3. Complete the Square: Factoring out 44 from the xx-terms, we get y=4(x2−4)y = 4(x^2 - 4). The term −4-4 inside the parentheses is not yet a perfect square.
  4. Add/Subtract Inside Parentheses: To complete the square, we need to add and subtract (b2a)2(\frac{b}{2a})^2 inside the parentheses, where aa is the coefficient of x2x^2 and bb is the coefficient of xx. Since there is no xx term, b=0b = 0, so (b2a)2=0(\frac{b}{2a})^2 = 0. Therefore, we do not need to add or subtract any term inside the parentheses.
  5. Equation in Vertex Form: The equation y=4(x2−4)y = 4(x^2 - 4) is already in the form y=a(x−h)2+ky = a(x - h)^2 + k, where a=4a = 4, h=0h = 0, and k=−16k = -16. Therefore, the vertex of the parabola is at (h,k)=(0,−16)(h, k) = (0, -16).

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