What is the range of this quadratic function? y = x^2 - 6x + 9 Choices: (A){y | y ≤ 0} (B){y | y ≥ 0} (C){y | y ≥ -3} (D)all real numbers

Identify the quadratic function: Identify the quadratic function. We are given the quadratic function y = x^2 - 6x + 9.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 = -6. x = -(-6)/(2*1) x = 6/2 x = 3Find vertex y-coordinate: Find the y-coordinate of the vertex by substituting x = 3 into the equation. y = (3)^2 - 6*(3) + 9 y = 9 - 18 + 9 y = 0Determine 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. The vertex of the parabola is (3, 0), 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 ≥ 0}

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? $\newline$ $y = x^2 - 6x + 9$ $\newline$ Choices: $\newline$ (A) ${y | y \leq 0}$ $\newline$ (B) ${y | y \geq 0}$ $\newline$ (C) ${y | y \geq -3}$ $\newline$ (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 - 6x + 9

\newline

Choices:

\newline

(A)

{y | y \leq 0}

\newline

(B)

{y | y \geq 0}

\newline

(C)

{y | y \geq -3}

\newline

(D)all real numbers

Identify the quadratic function: Identify the quadratic function. $\newline$ We are given the quadratic function $y = x^2 - 6x + 9$ .
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 $b = -6$ . $\newline$ $x = -(-6)/(2\cdot 1)$ $\newline$ $x = \frac{6}{2}$ $\newline$ $x = 3$
Find vertex y-coordinate: Find the y-coordinate of the vertex by substituting $x = 3$ into the equation. $y = (3)^2 - 6\times(3) + 9$ $y = 9 - 18 + 9$ $y = 0$
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 $(3, 0)$ , and since the parabola opens upwards, the $y$ -values must be greater than or equal to the $y$ -coordinate of the vertex. $\newline$ Range: \{ $y | y \geq 0$ \}