What is the range of this quadratic function? y = x^2 - 14x + 49 Choices: (A){y | y ≤ -7} (B){y | y ≥ -7} (C){y | y ≥ 0} (D)all real numbers

Find Vertex: We have the quadratic function y = x^2 - 14x + 49. 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.Calculate x-coordinate: Substitute a = 1 and b = -14 into the formula x = -b/(2a) to find the x-coordinate of the vertex. x = -(-14)/(2*1) x = 14/2 x = 7Calculate y-coordinate: Now we need to find the y-coordinate of the vertex by substituting x = 7 into the original equation y = x^2 - 14x + 49. y = (7)^2 - 14(7) + 49 y = 49 - 98 + 49 y = 0Determine Parabola Direction: The vertex of the quadratic function is (7, 0). Since the coefficient of x^2 is positive (a = 1), the parabola opens upwards. This means that the vertex represents the minimum point of the parabola.Find Range: Given that the parabola opens upwards and the y-coordinate of the vertex is 0, the range of the function is all y-values greater than or equal to the y-coordinate of the vertex. Range: {y | y ≥ 0}

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What is the range of this quadratic function? $\newline$ $y = x^2 - 14x + 49$ $\newline$ Choices: $\newline$ (A) $\{y | y \leq -7\}$ $\newline$ (B) $\{y | y \geq -7\}$ $\newline$ (C) $\{y | y \geq 0\}$ $\newline$ (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 - 14x + 49

\newline

Choices:

\newline

(A)

\{y | y \leq -7\}

\newline

(B)

\{y | y \geq -7\}

\newline

(C)

\{y | y \geq 0\}

\newline

(D)all real numbers

Find Vertex: We have the quadratic function $y = x^2 - 14x + 49$ . 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 $x$ .
Calculate x-coordinate: Substitute $a = 1$ and $b = -14$ into the formula $x = -\frac{b}{2a}$ to find the x-coordinate of the vertex. $\newline$ $x = -\frac{-14}{2\cdot 1}$ $\newline$ $x = \frac{14}{2}$ $\newline$ $x = 7$
Calculate y-coordinate: Now we need to find the y-coordinate of the vertex by substituting $x = 7$ into the original equation $y = x^2 - 14x + 49$ .
$y = (7)^2 - 14(7) + 49$
$y = 49 - 98 + 49$
$y = 0$
Determine Parabola Direction: The vertex of the quadratic function is $(7, 0)$ . Since the coefficient of $x^2$ is positive $(a = 1)$ , the parabola opens upwards. This means that the vertex represents the minimum point of the parabola.
Find Range: Given that the parabola opens upwards and the $y$ -coordinate of the vertex is $0$ , the range of the function is all $y$ -values greater than or equal to the $y$ -coordinate of the vertex. $\newline$ Range: $\{y \mid y \geq 0\}$