Determine the domain and the range of the function.
-$$f ( x ) = x ^ { 2 } - 2 x - 8$$
A) domain: all real numbers
range: all real numbers

B) domain: $\{ x \mid x \geq - 1 \}$
range: $\{ y \mid y \geq - 9 \}$

C) domain: all real numbers
range: $\{ y \mid y \geq - 9 \}$

D) domain: all real numbers
range: $\{ y \mid y \leq 9 \}$

Question

Determine the domain and the range of the function.
-$$f ( x ) = x ^ { 2 } - 2 x - 8$$
A) domain: all real numbers
range: all real numbers

B) domain: $\{ x \mid x \geq - 1 \}$  
range: $\{ y \mid y \geq - 9 \}$

C) domain: all real numbers
 range: $\{ y \mid y \geq - 9 \}$

D) domain: all real numbers
 range: $\{ y \mid y \leq 9 \}$

Quizplus · Accepted Answer

The function $f(x) = x^2 - 2x - 8$ is a quadratic function, which means its graph is a parabola. For any quadratic function in the form $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants and $a \neq 0$, the domain is all real numbers because you can plug any real number into the function and get a real number out. The range, however, depends on the direction of the parabola (upward or downward) and its vertex. Since the coefficient of $x^2$ is positive, the parabola opens upwards, and its minimum value is at the vertex. The vertex of a parabola given by $f(x) = ax^2 + bx + c$ can be found using the formula $-\frac{b}{2a}$ for the x-coordinate. Plugging the values from the given function, we get $-\frac{-2}{2(1)} = 1$ for the x-coordinate of the vertex. Substituting $x = 1$ back into the function gives the y-coordinate of the vertex, which is the minimum value of the function: $1^2 - 2(1) - 8 = -9$. Therefore, the range of the function is all real numbers greater than or equal to -9.

Determine the Domain and the Range of the Function f(x)=x2−2x−8f ( x ) = x ^ { 2 } - 2 x - 8f(x)=x2−2x−8

Determine the Domain and the Range of the Function $f ( x ) = x ^ { 2 } - 2 x - 8$