Find the range of the given function.
-$$f ( x ) = - \frac { 1 } { 3 } x ^ { 2 } + 3 x - 4$$
A) $\left[ \frac { 11 } { 4 } , \infty \right)$
B) $\left[ - \infty , - \frac { 97 } { 4 } \right]$
C) $[ - 4 , \infty )$
D) $\left( - \infty , \frac { 11 } { 4 } \right]$

Question

Quizplus · Accepted Answer

The given function is a quadratic function in the form of $f(x) = ax^2 + bx + c$ with $a = -\frac{1}{3}$, $b = 3$, and $c = -4$. Since $a < 0$, the parabola opens downwards, meaning it has a maximum point but no minimum point. To find the maximum value (which will be the highest value of the range), we use the vertex formula $x = -\frac{b}{2a}$. Substituting $a$ and $b$ gives $x = -\frac{3}{2(-1/3)} = \frac{3}{2/3} = \frac{9}{2}$. Substituting $x = \frac{9}{2}$ back into the original equation to find the maximum value $f(\frac{9}{2}) = -\frac{1}{3}(\frac{9}{2})^2 + 3(\frac{9}{2}) - 4$, which simplifies to $\frac{11}{4}$. Since the parabola opens downwards, the range is all real numbers less than or equal to this maximum value, hence $\left( - \infty , \frac { 11 } { 4 } \right]$.

Find the Range of the Given Function f(x)=−13x2+3x−4f ( x ) = - \frac { 1 } { 3 } x ^ { 2 } + 3 x - 4f(x)=−31​x2+3x−4

Find the Range of the Given Function $f ( x ) = - \frac { 1 } { 3 } x ^ { 2 } + 3 x - 4$