What is the range of this quadratic function? y = x^2 + 12x + 20 Choices: (A){y | y ≤ -6} (B){y | y ≥ -6} (C){y | y ≥ -16} (D)all real numbers

Calculate x-coordinate of vertex: We have the quadratic function y = x^2 + 12x + 20. To find the range, we need to determine the vertex of the parabola. The x-coordinate of the vertex is given by the formula x = -b/(2a). In our equation, a = 1 and b = 12. Calculate the x-coordinate of the vertex: x = -12/(2*1) x = -12/2 x = -6Calculate y-coordinate of vertex: Now that we have the x-coordinate of the vertex, we need to find the corresponding y-coordinate by substituting x = -6 into the original equation. Calculate the y-coordinate when x = -6: y = (-6)^2 + 12*(-6) + 20 y = 36 - 72 + 20 y = -36 + 20 y = -16Determine vertex of parabola: The vertex of the parabola is at the point (-6, -16). 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 of function: Since the parabola opens upwards and the vertex is the lowest point, the range of the function is all y-values greater than or equal to the y-coordinate of the vertex. The range of the function is {y | y ≥ -16}.

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What is the range of this quadratic function? $\newline$ $y = x^2 + 12x + 20$ $\newline$ Choices: $\newline$ (A) $\{y | y \leq -6\}$ $\newline$ (B) $\{y | y \geq -6\}$ $\newline$ (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 + 12x + 20

\newline

Choices:

\newline

(A)

\{y | y \leq -6\}

\newline

(B)

\{y | y \geq -6\}

\newline

(C)

\{y | y \geq -16\}

\newline

(D)all real numbers

Calculate x-coordinate of vertex: We have the quadratic function $y = x^2 + 12x + 20$ . To find the range, we need to determine the vertex of the parabola. The x-coordinate of the vertex is given by the formula $x = -\frac{b}{2a}$ . In our equation, $a = 1$ and $b = 12$ . $\newline$ Calculate the x-coordinate of the vertex: $\newline$ $x = -\frac{12}{2 \times 1}$ $\newline$ $x = -\frac{12}{2}$ $\newline$ $x = -6$
Calculate y-coordinate of vertex: Now that we have the x-coordinate of the vertex, we need to find the corresponding y-coordinate by substituting $x = -6$ into the original equation. $\newline$ Calculate the y-coordinate when $x = -6$ : $\newline$ $y = (-6)^2 + 12*(-6) + 20$ $\newline$ $y = 36 - 72 + 20$ $\newline$ $y = -36 + 20$ $\newline$ $y = -16$
Determine vertex of parabola: The vertex of the parabola is at the point $(-6, -16)$ . 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 of function: Since the parabola opens upwards and the vertex is the lowest point, the range of the function is all $y$ -values greater than or equal to the $y$ -coordinate of the vertex. $\newline$ The range of the function is $\{y \mid y \geq -16\}$ .