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What is the range of this quadratic function?\newliney=x2+2x−35y = x^2 + 2x - 35\newlineChoices:\newline(A){y∣y≤−1}\{y | y \leq -1\}\newline(B){y∣y≥−1}\{y | y \geq -1\}\newline(C){y∣y≥−36}\{y | y \geq -36\}\newline(D)all real numbers

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Q. What is the range of this quadratic function?\newliney=x2+2x−35y = x^2 + 2x - 35\newlineChoices:\newline(A){y∣y≤−1}\{y | y \leq -1\}\newline(B){y∣y≥−1}\{y | y \geq -1\}\newline(C){y∣y≥−36}\{y | y \geq -36\}\newline(D)all real numbers
  1. Find Vertex: We have the quadratic function y=x2+2x−35y = x^2 + 2x - 35. To find the range, we need to determine the vertex of the parabola.\newlineThe x-coordinate of the vertex is given by x=−b2ax = -\frac{b}{2a}. In our case, a=1a = 1 and b=2b = 2.\newlineSo, x=−22⋅1=−22=−1x = -\frac{2}{2\cdot1} = -\frac{2}{2} = -1.
  2. Calculate Vertex Coordinates: Now we need to find the y-coordinate of the vertex by substituting x=−1x = -1 into the equation y=x2+2x−35y = x^2 + 2x - 35.y=(−1)2+2(−1)−35=1−2−35=−36y = (-1)^2 + 2(-1) - 35 = 1 - 2 - 35 = -36.So, the vertex of the parabola is at the point (−1,−36)(-1, -36).
  3. Determine Parabola Direction: Since the coefficient of x2x^2 is positive (a=1a = 1), the parabola opens upwards. This means that the vertex represents the minimum point on the graph of the quadratic function.
  4. Identify Range: The range of the function is all the y-values that the function can take. Since the parabola opens upwards and the vertex is the lowest point, the range is all y-values greater than or equal to the y-coordinate of the vertex.\newlineTherefore, the range is y∣y≥−36{y | y \geq -36}.

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