What is the range of this quadratic function? y = x^2 + 8x + 16 Choices: (A){y | y ≥ 0} (B){y | y ≤ 0} (C){y | y ≥ -4} (D)all real numbers

Identify Quadratic Function: Identify the quadratic function and its general form. The given quadratic function is y = x^2 + 8x + 16, which is in the general form y = ax^2 + bx + c.Find Vertex x-coordinate: Find the x-coordinate of the vertex of the parabola. The x-coordinate of the vertex can be found using the formula x = -b/(2a). Here, a = 1 and b = 8. x = -8/(2*1) x = -8/2 x = -4Find Vertex y-coordinate: Find the y-coordinate of the vertex by substituting x = -4 into the quadratic function. y = (-4)^2 + 8*(-4) + 16 y = 16 - 32 + 16 y = 0 The vertex of the parabola is at the point (-4, 0).Determine Parabola Direction: Determine the direction in which the parabola opens. Since the coefficient of x^2 is positive (a = 1), the parabola opens upwards.Find Range of Function: Find the range of the quadratic function based on the vertex and the direction of the parabola. The vertex is at (-4, 0) and the parabola opens upwards, which means all y-values are greater than or equal to the y-coordinate of the vertex. Therefore, the range of the function is {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) ${y | y \geq 0}$ $\newline$ (B) ${y | y \leq 0}$ $\newline$ (C) ${y | y \geq -4}$ $\newline$ (D)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 + 8x + 16

\newline

Choices:

\newline

(A)

{y | y \geq 0}

\newline

(B)

{y | y \leq 0}

\newline

(C)

{y | y \geq -4}

\newline

(D)all real numbers

Identify Quadratic Function: Identify the quadratic function and its general form. The given quadratic function is $y = x^2 + 8x + 16$ , which is in the general form $y = ax^2 + bx + c$ .
Find Vertex x-coordinate: Find the x-coordinate of the vertex of the parabola. $\newline$ The x-coordinate of the vertex can be found using the formula $x = -\frac{b}{2a}$ . Here, $a = 1$ and $b = 8$ . $\newline$ $x = -\frac{8}{2 \times 1}$ $\newline$ $x = -\frac{8}{2}$ $\newline$ $x = -4$
Find Vertex y-coordinate: Find the y-coordinate of the vertex by substituting $x = -4$ into the quadratic function. $y = (-4)^2 + 8*(-4) + 16$ $y = 16 - 32 + 16$ $y = 0$ The vertex of the parabola is at the point $(-4, 0)$ .
Determine Parabola Direction: Determine the direction in which the parabola opens. Since the coefficient of $x^2$ is positive ( $a = 1$ ), the parabola opens upwards.
Find Range of Function: Find the range of the quadratic function based on the vertex and the direction of the parabola. $\newline$ The vertex is at $(-4, 0)$ and the parabola opens upwards, which means all $y$ -values are greater than or equal to the $y$ -coordinate of the vertex. $\newline$ Therefore, the range of the function is \{ $y | y \geq 0$ \}.