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What is the range of this quadratic function?\newliney=x2+12x+20y = x^2 + 12x + 20\newlineChoices:\newline(A){y∣y≤−6}\{y | y \leq -6\}\newline(B){y∣y≥−6}\{y | y \geq -6\}\newline(C){y∣y≥−16}\{y | y \geq -16\}\newline(D)all real numbers

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Q. What is the range of this quadratic function?\newliney=x2+12x+20y = x^2 + 12x + 20\newlineChoices:\newline(A){y∣y≤−6}\{y | y \leq -6\}\newline(B){y∣y≥−6}\{y | y \geq -6\}\newline(C){y∣y≥−16}\{y | y \geq -16\}\newline(D)all real numbers
  1. Calculate x-coordinate of vertex: We have the quadratic function y=x2+12x+20y = x^2 + 12x + 20. To find the range, we need to determine the vertex of the parabola. The x-coordinate of the vertex is given by the formula x=−b2ax = -\frac{b}{2a}. In our equation, a=1a = 1 and b=12b = 12.\newlineCalculate the x-coordinate of the vertex:\newlinex=−122×1x = -\frac{12}{2 \times 1}\newlinex=−122x = -\frac{12}{2}\newlinex=−6x = -6
  2. Calculate y-coordinate of vertex: Now that we have the x-coordinate of the vertex, we need to find the corresponding y-coordinate by substituting x=−6x = -6 into the original equation.\newlineCalculate the y-coordinate when x=−6x = -6:\newliney=(−6)2+12∗(−6)+20y = (-6)^2 + 12*(-6) + 20\newliney=36−72+20y = 36 - 72 + 20\newliney=−36+20y = -36 + 20\newliney=−16y = -16
  3. Determine vertex of parabola: The vertex of the parabola is at the point (−6,−16)(-6, -16). Since the coefficient of x2x^2 is positive (a=1a = 1), the parabola opens upwards. This means that the vertex represents the minimum point of the parabola.
  4. Find range of function: Since the parabola opens upwards and the vertex is the lowest point, the range of the function is all yy-values greater than or equal to the yy-coordinate of the vertex.\newlineThe range of the function is {y∣y≥−16}\{y \mid y \geq -16\}.

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