What is the range of this quadratic function? y = x^2 - 4x - 12 Choices: (A){y | y ≥ -16} (B){y | y ≥ 2} (C){y | y ≥ 16} (D)all real numbers

Find Vertex: We have the quadratic function y = x^2 - 4x - 12. 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 = -4 into the formula. x = -(-4)/(2*1) x = 4/2 x = 2Calculate y-coordinate: Now that we have the x-coordinate of the vertex, we need to find the corresponding y-coordinate. We do this by substituting x = 2 into the original equation. y = (2)^2 - 4(2) - 12 y = 4 - 8 - 12 y = -16Determine Parabola Direction: The vertex of the parabola is at the point (2, -16). Since the coefficient of x^2 is positive (a = 1), the parabola opens upwards. This means that the vertex represents the minimum point on the graph of the quadratic function.Identify Range: Since the parabola opens upwards and the y-coordinate of the vertex is the minimum value that y can take, the range of the function is all y-values greater than or equal to the y-coordinate of the vertex. Range: {y | y ≥ -16}

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What is the range of this quadratic function? $\newline$ $y = x^2 - 4x - 12$ $\newline$ Choices: $\newline$ (A) ${y | y \geq -16}$ $\newline$ (B) ${y | y \geq 2}$ $\newline$ (C) ${y | y \geq 16}$ $\newline$ (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 - 4x - 12

\newline

Choices:

\newline

(A)

{y | y \geq -16}

\newline

(B)

{y | y \geq 2}

\newline

(C)

{y | y \geq 16}

\newline

(D)all real numbers

Find Vertex: We have the quadratic function $y = x^2 - 4x - 12$ . 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 $b = -4$ into the formula. $\newline$ $x = -(-4)/(2\cdot1)$ $\newline$ $x$ $0$ $\newline$ $x$ $1$
Calculate y-coordinate: Now that we have the x-coordinate of the vertex, we need to find the corresponding y-coordinate. We do this by substituting $x = 2$ into the original equation. $\newline$ $y = (2)^2 - 4(2) - 12$ $\newline$ $y = 4 - 8 - 12$ $\newline$ $y = -16$
Determine Parabola Direction: The vertex of the parabola is at the point $(2, -16)$ . Since the coefficient of $x^2$ is positive $(a = 1)$ , the parabola opens upwards. This means that the vertex represents the minimum point on the graph of the quadratic function.
Identify Range: Since the parabola opens upwards and the $y$ -coordinate of the vertex is the minimum value that $y$ can take, the range of the function is all $y$ -values greater than or equal to the $y$ -coordinate of the vertex. $\newline$ Range: $\{y | y \geq -16\}$