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What is the range of this quadratic function?\newliney=x212x+32y = x^2 - 12x + 32\newlineChoices:\newline(A)yy4{y | y \geq -4}\newline(B)yy4{y | y \geq 4}\newline(C)yy6{y | y \geq 6}\newline(D)all real numbers

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Q. What is the range of this quadratic function?\newliney=x212x+32y = x^2 - 12x + 32\newlineChoices:\newline(A)yy4{y | y \geq -4}\newline(B)yy4{y | y \geq 4}\newline(C)yy6{y | y \geq 6}\newline(D)all real numbers
  1. Identify Quadratic Function: Identify the quadratic function.\newlineWe have the quadratic function y=x212x+32y = x^2 - 12x + 32.
  2. Find Vertex x-coordinate: Find the x-coordinate of the vertex.\newlineThe x-coordinate of the vertex of a quadratic function in the form y=ax2+bx+cy = ax^2 + bx + c is given by x=b2ax = -\frac{b}{2a}. Here, a=1a = 1 and b=12b = -12.\newlinex=(12)/(21)x = -(-12)/(2\cdot 1)\newlinex=122x = \frac{12}{2}\newlinex=6x = 6
  3. Find Vertex y-coordinate: Find the y-coordinate of the vertex.\newlineSubstitute x=6x = 6 into the quadratic function to find the y-coordinate.\newliney=(6)212(6)+32y = (6)^2 - 12(6) + 32\newliney=3672+32y = 36 - 72 + 32\newliney=36+32y = -36 + 32\newliney=4y = -4
  4. Determine Parabola Direction: Determine the direction of the parabola. Since a=1a = 1 and a>0a > 0, the parabola opens upwards.
  5. Find Range of Function: Find the range of the function.\newlineThe vertex of the parabola is (6,4)(6, -4), and since the parabola opens upwards, the yy-values must be greater than or equal to the yy-coordinate of the vertex.\newlineRange: \{yy4y | y \geq -4\}

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