Determine the domain and the range of the function.
-$$f ( x ) = - 11 x ^ { 2 } - 2 x - 8$$
A) domain: all real numbers
range: $\left\{ \mathrm { y } \mid \mathrm { y } \leq - \frac { 87 } { 11 } \right\}$
B) domain: all real numbers
range: $\left\{ y \mid y \geq \frac { 87 } { 11 } \right\}$
C) domain: all real numbers
range: $\left\{ y \mid y \leq \frac { 87 } { 11 } \right\}$
D) domain: all real numbers
range: $\left\{ y \mid y \geq - \frac { 87 } { 11 } \right\}$

Question

Quizplus · Accepted Answer

The function $f(x) = -11x^2 - 2x - 8$ is a quadratic function, which means its graph is a parabola. Since the coefficient of $x^2$ is negative (-11), the parabola opens downwards. This implies that the function has a maximum value but no minimum value. The domain of any quadratic function is all real numbers because you can input any real number for $x$. To find the maximum value (which affects the range), we can use the vertex formula. The vertex of a parabola given by $f(x) = ax^2 + bx + c$ is at $(-\frac{b}{2a}, f(-\frac{b}{2a}))$. For this function, $a = -11$, $b = -2$, so the x-coordinate of the vertex is $-\frac{-2}{2(-11)} = \frac{1}{11}$. Plugging this back into the function gives the y-coordinate (maximum value) of the vertex as $f(\frac{1}{11}) = -11(\frac{1}{11})^2 - 2(\frac{1}{11}) - 8 = -\frac{87}{11}$. Therefore, the range is all real numbers less than or equal to $-\frac{87}{11}$, since the parabola opens downwards and this is its maximum value.

Determine the Domain and the Range of the Function f(x)=−11x2−2x−8f ( x ) = - 11 x ^ { 2 } - 2 x - 8f(x)=−11x2−2x−8

Determine the Domain and the Range of the Function $f ( x ) = - 11 x ^ { 2 } - 2 x - 8$