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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}\left\{y\mid y \geq 0\right\}\newline(B) {y∣y≤0}\left\{y\mid y \leq 0\right\}\newline(C) {y∣y≥−4}\left\{y\mid y \geq -4\right\}\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}\left\{y\mid y \geq 0\right\}\newline(B) {y∣y≤0}\left\{y\mid y \leq 0\right\}\newline(C) {y∣y≥−4}\left\{y\mid y \geq -4\right\}\newline(D) all real numbers
  1. Identify general form: Identify the general form of the quadratic function.\newlineThe given function is y=x2+8x+16y = x^2 + 8x + 16, which is in the standard form y=ax2+bx+cy = ax^2 + bx + c.
  2. Find vertex x-coordinate: Find the x-coordinate of the vertex.\newlineThe x-coordinate of the vertex of a parabola in the form y=ax2+bx+cy = ax^2 + bx + c is given by −b2a-\frac{b}{2a}. Here, a=1a = 1 and b=8b = 8.\newlinex=−82⋅1=−82=−4x = -\frac{8}{2 \cdot 1} = -\frac{8}{2} = -4.
  3. Find vertex y-coordinate: Find the y-coordinate of the vertex by substituting x=−4x = -4 into the equation.y=(−4)2+8(−4)+16y = (-4)^2 + 8(-4) + 16y=16−32+16y = 16 - 32 + 16y=0y = 0
  4. Determine parabola direction: Determine the direction in which the parabola opens.\newlineSince a=1a = 1 and a>0a > 0, the parabola opens upwards.
  5. Find range of function: Find the range of the function based on the vertex and the direction of the parabola.\newlineThe vertex is (−4,0)(-4, 0), and since the parabola opens upwards, the range of yy values starts at the yy-coordinate of the vertex and goes to infinity.\newlineRange: {y\{ y | y≥0}y \geq 0 \}

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