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What is the range of this quadratic function? $\newline$ $y = x^2 + 2x + 1$ $\newline$ Choices: $\newline$ (A) ${y | y \geq 1}$ $\newline$ (B) ${y | y \geq 0}$ $\newline$ (C) ${y | y \leq 0}$ $\newline$ (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 + 2x + 1

\newline

Choices:

\newline

(A)

{y | y \geq 1}

\newline

(B)

{y | y \geq 0}

\newline

(C)

{y | y \leq 0}

\newline

(D)all real numbers

Identify the quadratic function: Identify the quadratic function. $\newline$ We have the quadratic function $y = x^2 + 2x + 1$ .
Find vertex x-coordinate: Find the x-coordinate of the vertex. $\newline$ The general form of a quadratic function is $y = ax^2 + bx + c$ . To find the x-coordinate of the vertex, use the formula $x = -\frac{b}{2a}$ . $\newline$ For our function, $a = 1$ and $b = 2$ . $\newline$ $x = -\frac{2}{2\cdot 1}$ $\newline$ $x = -\frac{2}{2}$ $\newline$ $x = -1$
Find vertex y-coordinate: Find the y-coordinate of the vertex by substituting $x = -1$ into the function. $\newline$ $y = (-1)^2 + 2(-1) + 1$ $\newline$ $y = 1 - 2 + 1$ $\newline$ $y = 0$
Determine parabola direction: Determine the direction in which the parabola opens. Since $a = 1$ and $a > 0$ , the parabola opens upwards.
Find range: Find the range of the function. $\newline$ The vertex of the parabola is $(-1, 0)$ , and since the parabola opens upwards, the $y$ -values 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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