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What is the range of this quadratic function?\newliney=x2+2x−15y = x^2 + 2x - 15\newlineChoices:\newline(A)y∣y≥−16{y | y \geq -16}\newline(B)y∣y≤−16{y | y \leq -16}\newline(C)y∣y≥−1{y | y \geq -1}\newline(D)all real numbers

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Q. What is the range of this quadratic function?\newliney=x2+2x−15y = x^2 + 2x - 15\newlineChoices:\newline(A)y∣y≥−16{y | y \geq -16}\newline(B)y∣y≤−16{y | y \leq -16}\newline(C)y∣y≥−1{y | y \geq -1}\newline(D)all real numbers
  1. Identify Quadratic Function: Identify the quadratic function and its coefficients.\newlineThe given quadratic function is y=x2+2x−15y = x^2 + 2x - 15. The coefficients are a=1a = 1, b=2b = 2, and c=−15c = -15.
  2. Find Vertex x-coordinate: Find the x-coordinate of the vertex.\newlineThe x-coordinate of the vertex of a parabola given by y=ax2+bx+cy = ax^2 + bx + c is found using the formula x=−b2ax = -\frac{b}{2a}. Here, a=1a = 1 and b=2b = 2.\newlinex=−22⋅1x = -\frac{2}{2\cdot 1}\newlinex=−22x = -\frac{2}{2}\newlinex=−1x = -1
  3. Find Vertex y-coordinate: Find the y-coordinate of the vertex by substituting x=−1x = -1 into the quadratic function.y=(−1)2+2(−1)−15y = (-1)^2 + 2(-1) - 15y=1−2−15y = 1 - 2 - 15y=−16y = -16
  4. Determine Parabola Direction: Determine the direction in which the parabola opens.\newlineSince the coefficient a=1a = 1 is positive, the parabola opens upwards.
  5. Find Range: Find the range of the quadratic function.\newlineThe vertex of the parabola is (−1,−16)(-1, -16), and since the parabola opens upwards, the yy-values will be greater than or equal to the yy-coordinate of the vertex.\newlineRange: \{y∣y≥−16y | y \geq -16\}

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