What is the range of this quadratic function? y = x^2 + 16x + 64 Choices: (A){y | y ≤ 0} (B){y | y ≥ 0} (C){y | y ≥ -8} (D)all real numbers

Find Vertex: We have the quadratic function y = x^2 + 16x + 64. To find the range, we need to determine the vertex of the parabola. The x-coordinate of the vertex can be found using the formula x = -b/(2a), where a is the coefficient of x^2 and b is the coefficient of x. Substitute a = 1 and b = 16 into the formula. x = -16/(2*1) x = -16/2 x = -8Calculate Vertex Coordinates: Now that we have the x-coordinate of the vertex, we need to find the corresponding y-coordinate by substituting x = -8 into the original equation. y = (-8)^2 + 16*(-8) + 64 y = 64 - 128 + 64 y = 0 The vertex of the parabola is at the point (-8, 0).Determine Parabola Direction: The coefficient of x^2 in the equation y = x^2 + 16x + 64 is positive, which means the parabola opens upwards. Since the parabola opens upwards and the vertex is the lowest point on the graph, the range of the function will include all y-values greater than or equal to the y-coordinate of the vertex.Identify Range: The y-coordinate of the vertex is 0, so the range of the function is all y-values greater than or equal to 0. Therefore, the range of the function is {y | y ≥ 0}.

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What is the range of this quadratic function? $\newline$ $y = x^2 + 16x + 64$ $\newline$ Choices: $\newline$ (A) ${y | y \leq 0}$ $\newline$ (B) ${y | y \geq 0}$ $\newline$ (C) ${y | y \geq -8}$ $\newline$ (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 + 16x + 64

\newline

Choices:

\newline

(A)

{y | y \leq 0}

\newline

(B)

{y | y \geq 0}

\newline

(C)

{y | y \geq -8}

\newline

(D)all real numbers

Find Vertex: We have the quadratic function $y = x^2 + 16x + 64$ . To find the range, we need to determine the vertex of the parabola. The $x$ -coordinate of the vertex can be found using the formula $x = -\frac{b}{2a}$ , where $a$ is the coefficient of $x^2$ and $b$ is the coefficient of $x$ . $\newline$ Substitute $a = 1$ and $b = 16$ into the formula. $\newline$ $x = -\frac{16}{2 \cdot 1}$ $\newline$ $x$ $0$ $\newline$ $x$ $1$
Calculate Vertex Coordinates: Now that we have the x-coordinate of the vertex, we need to find the corresponding y-coordinate by substituting $x = -8$ into the original equation. $y = (-8)^2 + 16*(-8) + 64$ $y = 64 - 128 + 64$ $y = 0$ The vertex of the parabola is at the point $(-8, 0)$ .
Determine Parabola Direction: The coefficient of $x^2$ in the equation $y = x^2 + 16x + 64$ is positive, which means the parabola opens upwards. Since the parabola opens upwards and the vertex is the lowest point on the graph, the range of the function will include all $y$ -values greater than or equal to the $y$ -coordinate of the vertex.
Identify Range: The $y$ -coordinate of the vertex is $0$ , so the range of the function is all $y$ -values greater than or equal to $0$ . Therefore, the range of the function is $\{y | y \geq 0\}$ .