What is the range of this quadratic function? y = x^2 - 12x + 32 Choices: (A){y | y ≥ -4} (B){y | y ≥ 4} (C){y | y ≥ 6} (D)all real numbers

Identify Quadratic Function: Identify the quadratic function. We have the quadratic function y = x^2 - 12x + 32.Find Vertex x-coordinate: Find the x-coordinate of the vertex. The x-coordinate of the vertex of a quadratic function in the form y = ax^2 + bx + c is given by x = -b/(2a). Here, a = 1 and b = -12. x = -(-12)/(2*1) x = 12/2 x = 6Find Vertex y-coordinate: Find the y-coordinate of the vertex. Substitute x = 6 into the quadratic function to find the y-coordinate. y = (6)^2 - 12(6) + 32 y = 36 - 72 + 32 y = -36 + 32 y = -4Determine 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. The vertex of the parabola is (6, -4), and since the parabola opens upwards, the y-values must be greater than or equal to the y-coordinate of the vertex. Range: {y | y ≥ -4}

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

What is the range of this quadratic function? $\newline$ $y = x^2 - 12x + 32$ $\newline$ Choices: $\newline$ (A) ${y | y \geq -4}$ $\newline$ (B) ${y | y \geq 4}$ $\newline$ (C) ${y | y \geq 6}$ $\newline$ (D)all real numbers

View step-by-step help

Home

Math Problems

Algebra 1

Domain and range of quadratic functions: equations

Full solution

Q. What is the range of this quadratic function?

\newline

y = x^2 - 12x + 32

\newline

Choices:

\newline

(A)

{y | y \geq -4}

\newline

(B)

{y | y \geq 4}

\newline

(C)

{y | y \geq 6}

\newline

(D)all real numbers

Identify Quadratic Function: Identify the quadratic function. $\newline$ We have the quadratic function $y = x^2 - 12x + 32$ .
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 y-coordinate. $\newline$ $y = (6)^2 - 12(6) + 32$ $\newline$ $y = 36 - 72 + 32$ $\newline$ $y = -36 + 32$ $\newline$ $y = -4$
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 $(6, -4)$ , and since the parabola opens upwards, the $y$ -values must be greater than or equal to the $y$ -coordinate of the vertex. $\newline$ Range: \{ $y | y \geq -4$ \}