What is the range of this quadratic function? y = x^2 + 8x + 16 Choices: [[{y|y ≥ 0}][{y|y ≤ 0}][{y|y ≥ –4}][all real numbers]]

Identify general form: Identify the general form of the quadratic function. The given function is y = x^2 + 8x + 16, which is in the standard form y = ax^2 + bx + c.Find vertex x-coordinate: Find the x-coordinate of the vertex. The x-coordinate of the vertex of a parabola in the form y = ax^2 + bx + c is given by -b/(2a). Here, a = 1 and b = 8. x = -8/(2*1) = -8/2 = -4.Find vertex y-coordinate: Find the y-coordinate of the vertex by substituting x = -4 into the equation. y = (-4)^2 + 8*(-4) + 16 y = 16 - 32 + 16 y = 0Determine parabola direction: Determine the direction in which the parabola opens. Since a = 1 and a > 0, the parabola opens upwards.Find range of function: Find the range of the function based on the vertex and the direction of the parabola. The vertex is (-4, 0), and since the parabola opens upwards, the range of y values starts at the y-coordinate of the vertex and goes to infinity. Range: {y|y ≥ 0}

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What is the range of this quadratic function? $\newline$ $y = x^2 + 8x + 16$ $\newline$ Choices: $\newline$ (A) $\left\{y\mid y \geq 0\right\}$ $\newline$ (B) $\left\{y\mid y \leq 0\right\}$ $\newline$ (C) $\left\{y\mid y \geq -4\right\}$ $\newline$ (D) all real numbers

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Algebra 2

Domain and range of quadratic functions: equations

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Q. What is the range of this quadratic function?

\newline

y = x^2 + 8x + 16

\newline

Choices:

\newline

(A)

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

\newline

(B)

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

\newline

(C)

\left\{y\mid y \geq -4\right\}

\newline

(D) all real numbers

Identify general form: Identify the general form of the quadratic function. $\newline$ The given function is $y = x^2 + 8x + 16$ , which is in the standard form $y = ax^2 + bx + c$ .
Find vertex x-coordinate: Find the x-coordinate of the vertex. $\newline$ The x-coordinate of the vertex of a parabola in the form $y = ax^2 + bx + c$ is given by $-\frac{b}{2a}$ . Here, $a = 1$ and $b = 8$ . $\newline$ $x = -\frac{8}{2 \cdot 1} = -4$ .
Find vertex y-coordinate: Find the y-coordinate of the vertex by substituting $x = -4$ into the equation. $y = (-4)^2 + 8(-4) + 16$ $y = 16 - 32 + 16$ $y = 0$
Determine parabola direction: Determine the direction in which the parabola opens. $\newline$ Since $a = 1$ and a > 0, the parabola opens upwards.
Find range of function: Find the range of the function based on the vertex and the direction of the parabola. $\newline$ The vertex is $(-4, 0)$ , and since the parabola opens upwards, the range of $y$ values starts at the $y$ -coordinate of the vertex and goes to infinity. $\newline$ Range: $\{ y$ | $y \geq 0 \}$