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What is the range of this quadratic function? $\newline$ $y = x^2 - 12x + 35$ $\newline$ Choices: $\newline$ (A) ${y | y \geq 6}$ $\newline$ (B) ${y | y \geq -1}$ $\newline$ (C) ${y | y \leq -1}$ $\newline$ (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 - 12x + 35

\newline

Choices:

\newline

(A)

{y | y \geq 6}

\newline

(B)

{y | y \geq -1}

\newline

(C)

{y | y \leq -1}

\newline

(D)all real numbers

Find Vertex: We have the quadratic function $y = x^2 - 12x + 35$ . To find the range, we first 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 $x$ .
Calculate x-coordinate: Substitute $a = 1$ and $b = -12$ into the formula $x = -\frac{b}{2a}$ to find the x-coordinate of the vertex. $\newline$ $x = -\frac{-12}{2 \times 1}$ $\newline$ $x = \frac{12}{2}$ $\newline$ $x = 6$
Calculate y-coordinate: Now that we have the x-coordinate of the vertex, we can find the y-coordinate by substituting $x = 6$ into the original equation $y = x^2 - 12x + 35$ .
$y = (6)^2 - 12(6) + 35$
$y = 36 - 72 + 35$
$y = -36 + 35$
$y = -1$
Determine Parabola Direction: The vertex of the parabola is $(6, -1)$ . Since the coefficient of $x^2$ is positive $(a = 1)$ , the parabola opens upwards. This means that the vertex represents the minimum point on the graph of the quadratic function.
Find Range: Given that the parabola opens upwards and the y-coordinate of the vertex is $-1$ , the range of the function is all y-values that are greater than or equal to $-1$ . $Range: \{y | y \geq -1\}$

More problems from Domain and range of quadratic functions: equations

Question

h

is defined over the real numbers. This table gives a few values of

h

\newline

\begin{tabular}{lllll}

\newline

x

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1

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01

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9

\newline

h(x)

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1

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