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What is the range of this quadratic function?\newliney=x2−4x+4y = x^2 - 4x + 4\newlineChoices:\newline(A)y∣y≥2{y | y \geq 2}\newline(B)y∣y≤0{y | y \leq 0}\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=x2−4x+4y = x^2 - 4x + 4\newlineChoices:\newline(A)y∣y≥2{y | y \geq 2}\newline(B)y∣y≤0{y | y \leq 0}\newline(C)y∣y≥0{y | y \geq 0}\newline(D)all real numbers
  1. Identify Quadratic Function: Identify the general form of the quadratic function.\newlineThe given function is y=x2−4x+4y = x^2 - 4x + 4, which is in the standard form y=ax2+bx+cy = ax^2 + bx + c, where a=1a = 1, b=−4b = -4, and c=4c = 4.
  2. Find Vertex X-coordinate: Find the x-coordinate of the vertex of the parabola.\newlineThe x-coordinate of the vertex can be found using the formula x=−b2ax = -\frac{b}{2a}. Substituting the values of aa and bb, we get:\newlinex=−−42⋅1=42=2x = -\frac{-4}{2\cdot 1} = \frac{4}{2} = 2.
  3. Find Vertex Y-coordinate: Find the y-coordinate of the vertex by substituting the x-coordinate back into the original equation.\newlineSubstitute x=2x = 2 into y=x2−4x+4y = x^2 - 4x + 4 to find the y-coordinate:\newliney=(2)2−4(2)+4=4−8+4=0y = (2)^2 - 4(2) + 4 = 4 - 8 + 4 = 0.
  4. Determine Parabola Direction: Determine the direction in which the parabola opens.\newlineSince the coefficient of x2x^2 (a=1a = 1) is positive, the parabola opens upwards.
  5. Determine Range: Determine the range of the quadratic function.\newlineGiven that the parabola opens upwards and the vertex is at (2,0)(2, 0), the lowest point on the parabola is at y=0y = 0. Therefore, the range of the function is all yy-values greater than or equal to 00.

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