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

\newline

Choices:

\newline

(A)

{y | y \geq 0}

\newline

(B)

{y | y \leq 0}

\newline

(C)

{y | y \geq -7}

\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 $x$ . $\newline$ Substitute $a = 1$ and $b = 14$ into the formula. $\newline$ $x = -\frac{14}{2 \cdot 1}$ $\newline$ $x$ $0$ $\newline$ $x$ $1$
Calculate Y-coordinate: Now that we have the x-coordinate of the vertex, we need to find the y-coordinate by substituting $x = -7$ into the original equation. $y = (-7)^2 + 14*(-7) + 49$ $y = 49 - 98 + 49$ $y = 0$
Determine Parabola Direction: We have found the vertex of the parabola to be $(-7, 0)$ . Now we need to determine the direction in which the parabola opens. Since the coefficient of $x^2$ $(a = 1)$ is positive, the parabola opens upwards.
Identify Range: With the vertex at $(-7, 0)$ and the parabola opening upwards, the lowest point on the graph is the vertex. This means that the $y$ -values of the function must be greater than or equal to the $y$ -coordinate of the vertex. $\newline$ Therefore, the range of the function is \{ $y$ | $y \geq 0$ \}.

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