Determine where the function is increasing and where it is decreasing.
-$$g ( x ) = 6 x ^ { 2 } + 120 x + 564$$
A) increasing on $( - \infty , - 10 )$
decreasing on $( - 10 , \infty )$
B) decreasing on $( - \infty , 10 )$
increasing on $( 10 , \infty )$
C) increasing on $( - \infty , - 60 )$
decreasing on $( - 60 , \infty )$
D) decreasing on $( - \infty , - 10 )$
increasing on $( - 10 , \infty )$

Question

Quizplus · Accepted Answer

The function $g(x) = 6x^2 + 120x + 564$ is a quadratic function with a positive leading coefficient (6), which means it opens upwards and has a minimum point. The vertex of a quadratic function given in the form $ax^2 + bx + c$ can be found using $-\frac{b}{2a}$. Here, $a = 6$ and $b = 120$, so the vertex is at $-\frac{120}{2*6} = -\frac{120}{12} = -10$. Since the parabola opens upwards, the function is decreasing on the left side of the vertex and increasing on the right side of the vertex. Therefore, it is decreasing on $(- \infty, -10)$ and increasing on $(-10, \infty)$.

Determine Where the Function Is Increasing and Where It Is g(x)=6x2+120x+564g ( x ) = 6 x ^ { 2 } + 120 x + 564g(x)=6x2+120x+564

Determine Where the Function Is Increasing and Where It Is $g ( x ) = 6 x ^ { 2 } + 120 x + 564$