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What is the range of this quadratic function? $\newline$ $y = x^2 - 12x + 27$ $\newline$ Choices: $\newline$ (A) ${y | y \geq 9}$ $\newline$ (B) ${y | y \geq 6}$ $\newline$ (C) ${y | y \geq -9}$ $\newline$ (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 - 12x + 27

\newline

Choices:

\newline

(A)

{y | y \geq 9}

\newline

(B)

{y | y \geq 6}

\newline

(C)

{y | y \geq -9}

\newline

(D)all real numbers

Identify quadratic function: Identify the quadratic function. $\newline$ We have the quadratic function $y = x^2 - 12x + 27$ . We need to find the range of this function.
Find vertex x-coordinate: Find the x-coordinate of the vertex. $\newline$ The x-coordinate of the vertex of a quadratic function in the form $y = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$ . Here, $a = 1$ and $b = -12$ . $\newline$ $x = -(-12)/(2\cdot 1)$ $\newline$ $x = \frac{12}{2}$ $\newline$ $x = 6$
Find vertex y-coordinate: Find the y-coordinate of the vertex. $\newline$ Substitute $x = 6$ into the quadratic function to find the corresponding y-coordinate. $\newline$ $y = (6)^2 - 12(6) + 27$ $\newline$ $y = 36 - 72 + 27$ $\newline$ $y = -36 + 27$ $\newline$ $y = -9$
Determine parabola direction: Determine the direction of the parabola. Since $a = 1$ and $a > 0$ , the parabola opens upwards.
Find range of function: Find the range of the function. $\newline$ The vertex of the parabola is $(6, -9)$ , and since the parabola opens upwards, the $y$ -values must be greater than or equal to the $y$ -coordinate of the vertex. $\newline$ Range: \{ $y | y \geq -9$ \}

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