Find the range of the quadratic function.
-$$f ( x ) = - 5 x ^ { 2 } + 15 x$$
A) $\left( - \infty , \frac { 3 } { 2 } \right]$
B) $\left( - \infty , \frac { 45 } { 4 } \right]$
C) $\left( - \infty , - \frac { 3 } { 2 } \right]$
D) $\left( - \infty , - \frac { 45 } { 4 } \right]$

Question

Quizplus · Accepted Answer

The quadratic function $f(x) = -5x^2 + 15x$ opens downwards because the coefficient of $x^2$ is negative. To find the maximum value (vertex), we use the formula for the x-coordinate of the vertex of a quadratic function, $x = -\frac{b}{2a}$, where $a = -5$ and $b = 15$. Substituting these values gives $x = -\frac{15}{2(-5)} = \frac{3}{2}$. Substituting $x = \frac{3}{2}$ back into the function gives the maximum value $f\left(\frac{3}{2}\right) = -5\left(\frac{3}{2}\right)^2 + 15\left(\frac{3}{2}\right) = -\frac{45}{4} + \frac{45}{2} = \frac{45}{4}$. Therefore, the range is $\left(-\infty, \frac{45}{4}\right]$.

Find the Range of the Quadratic Function f(x)=−5x2+15xf ( x ) = - 5 x ^ { 2 } + 15 xf(x)=−5x2+15x

Find the Range of the Quadratic Function $f ( x ) = - 5 x ^ { 2 } + 15 x$