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What is the range of this quadratic function?\newliney=x2−14x+49y = x^2 - 14x + 49\newlineChoices:\newline(A){y∣y≤−7}\{y | y \leq -7\}\newline(B){y∣y≥−7}\{y | y \geq -7\}\newline(C){y∣y≥0}\{y | y \geq 0\}\newline(D)all real numbers

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Q. What is the range of this quadratic function?\newliney=x2−14x+49y = x^2 - 14x + 49\newlineChoices:\newline(A){y∣y≤−7}\{y | y \leq -7\}\newline(B){y∣y≥−7}\{y | y \geq -7\}\newline(C){y∣y≥0}\{y | y \geq 0\}\newline(D)all real numbers
  1. Find Vertex: We have the quadratic function y=x2−14x+49y = x^2 - 14x + 49. To find the range, we need to determine the vertex of the parabola. The xx-coordinate of the vertex can be found using the formula x=−b2ax = -\frac{b}{2a}, where aa is the coefficient of x2x^2 and bb is the coefficient of xx.
  2. Calculate x-coordinate: Substitute a=1a = 1 and b=−14b = -14 into the formula x=−b2ax = -\frac{b}{2a} to find the x-coordinate of the vertex.\newlinex=−−142⋅1x = -\frac{-14}{2\cdot 1}\newlinex=142x = \frac{14}{2}\newlinex=7x = 7
  3. Calculate y-coordinate: Now we need to find the y-coordinate of the vertex by substituting x=7x = 7 into the original equation y=x2−14x+49y = x^2 - 14x + 49.
    y=(7)2−14(7)+49y = (7)^2 - 14(7) + 49
    y=49−98+49y = 49 - 98 + 49
    y=0y = 0
  4. Determine Parabola Direction: The vertex of the quadratic function is (7,0)(7, 0). Since the coefficient of x2x^2 is positive (a=1)(a = 1), the parabola opens upwards. This means that the vertex represents the minimum point of the parabola.
  5. Find Range: Given that the parabola opens upwards and the yy-coordinate of the vertex is 00, the range of the function is all yy-values greater than or equal to the yy-coordinate of the vertex.\newlineRange: {y∣y≥0}\{y \mid y \geq 0\}

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