What is the range of this quadratic function? y = x^2 – 4x + 4 Choices: [[{y|y ≥ 2}][{y|y ≤ 0}][{y|y ≥ 0}][all real numbers]]

Calculate x-coordinate of vertex: Find the x-coordinate of the vertex using the formula x = -b/(2a). Here, a = 1 and b = -4. x = -(-4)/(2*1) = 4/2 = 2.Find y-coordinate of vertex: Plug x = 2 into the equation to find the y-coordinate of the vertex. y = (2)^2 - 4*(2) + 4 = 4 - 8 + 4 = 0.Determine vertex and parabola direction: The vertex is (2, 0). Since a = 1, which is positive, the parabola opens upwards.Identify range of parabola: Since the parabola opens upwards and the vertex has a y-coordinate of 0, the range is all y-values greater than or equal to 0. Range: {y|y ≥ 0}.

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What is the range of this quadratic function? $\newline$ $y = x^2 - 4x + 4$ $\newline$ Choices: $\newline$ $\left\{y \mid y \geq 2\right\}$ $\newline$ $\left\{y \mid y \leq 0\right\}$ $\newline$ $\left\{y \mid y \geq 0\right\}$ $\newline$ $\text{all real numbers}$

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Algebra 1

Domain and range of quadratic functions: equations

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Q. What is the range of this quadratic function?

\newline

y = x^2 - 4x + 4

\newline

Choices:

\newline

\left\{y \mid y \geq 2\right\}

\newline

\left\{y \mid y \leq 0\right\}

\newline

\left\{y \mid y \geq 0\right\}

\newline

\text{all real numbers}

Calculate x-coordinate of vertex: Find the x-coordinate of the vertex using the formula $x = -\frac{b}{2a}$ . Here, $a = 1$ and $b = -4$ . $x = -(-4)/(2\cdot 1) = \frac{4}{2} = 2$ .
Find y-coordinate of vertex: Plug $x = 2$ into the equation to find the y-coordinate of the vertex. $y = (2)^2 - 4\times(2) + 4 = 4 - 8 + 4 = 0.$
Determine vertex and parabola direction: The vertex is $(2, 0)$ . Since $a = 1$ , which is positive, the parabola opens upwards.
Identify range of parabola: Since the parabola opens upwards and the vertex has a y-coordinate of $0$ , the range is all y-values greater than or equal to $0$ . $\newline$ Range: $\{y|y \geq 0\}$ .