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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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Domain and range of quadratic functions: equations

Full solution

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 $\{y | y \geq 0\}$ .

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