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What is the range of this quadratic function?\newliney=x2+8x+16y = x^2 + 8x + 16\newlineChoices:\newline(A)y∣y≥0{y | y \geq 0}\newline(B)y∣y≤0{y | y \leq 0}\newline(C)y∣y≥−4{y | y \geq -4}\newline(D)all real numbers

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Q. What is the range of this quadratic function?\newliney=x2+8x+16y = x^2 + 8x + 16\newlineChoices:\newline(A)y∣y≥0{y | y \geq 0}\newline(B)y∣y≤0{y | y \leq 0}\newline(C)y∣y≥−4{y | y \geq -4}\newline(D)all real numbers
  1. Identify Quadratic Function: Identify the quadratic function and its general form. The given quadratic function is y=x2+8x+16y = x^2 + 8x + 16, which is in the general form y=ax2+bx+cy = ax^2 + bx + c.
  2. Find Vertex x-coordinate: Find the x-coordinate of the vertex of the parabola.\newlineThe x-coordinate of the vertex can be found using the formula x=−b2ax = -\frac{b}{2a}. Here, a=1a = 1 and b=8b = 8.\newlinex=−82×1x = -\frac{8}{2 \times 1}\newlinex=−82x = -\frac{8}{2}\newlinex=−4x = -4
  3. Find Vertex y-coordinate: Find the y-coordinate of the vertex by substituting x=−4x = -4 into the quadratic function.y=(−4)2+8∗(−4)+16y = (-4)^2 + 8*(-4) + 16y=16−32+16y = 16 - 32 + 16y=0y = 0The vertex of the parabola is at the point (−4,0)(-4, 0).
  4. Determine Parabola Direction: Determine the direction in which the parabola opens. Since the coefficient of x2x^2 is positive (a=1a = 1), the parabola opens upwards.
  5. Find Range of Function: Find the range of the quadratic function based on the vertex and the direction of the parabola.\newlineThe vertex is at (−4,0)(-4, 0) and the parabola opens upwards, which means all yy-values are greater than or equal to the yy-coordinate of the vertex.\newlineTherefore, the range of the function is \{y∣y≥0y | y \geq 0\}.

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