An absolute value function with a vertex of (4,14) is negative on the interval (−oo,−2). On what other interval is the function negative?

Step 1
The absolute value function with vertex (h,k) is defined as f(x)=a ${\displaystyle \left|x-h\right|+k}$.
The vertex is (4,14).
Therefore, ${\displaystyle f\left(x\right)=\left|x-4\right|+14}$.
The function can be written without absolute value as ${\displaystyle f\left(x\right)=\{\begin{array}{cc}a(x-4)+14& x>4\\ -a(x-4)+14& x<4\end{array}}$
Step 2
The function is negative on the interval ${\displaystyle (-\mathrm{\infty },-2)}$. Since the function is continuous, the value of the function at $x=-2$ will be zero.
Therefore,
$-a(-2-4)+14=0$
$-a(-6)+14=0$
$6a+14=0$
$6a=-14$
${\displaystyle a=\frac{-14}{6}}$
${\displaystyle =-\frac{7}{3}}$
Step 3
Thus, the function becomes ${\displaystyle f\left(x\right)=-\frac{7}{3}\left|x-4\right|+14}$.
Find the intervals on which ${\displaystyle f\left(x\right)=-\frac{7}{3}\left|x-4\right|+14}$ is negative as follows.
${\displaystyle -\frac{7}{3}\left|x-4\right|+14<0}$
${\displaystyle -\frac{7}{3}\left|x-4\right|<-14}$
${\displaystyle -\frac{7}{3}\left|x-4\right|>14}$
${\displaystyle 3\cdot \frac{7}{3}\left|x-4\right|>14\cdot 3}$
${\displaystyle 7\left|x-4\right|>42}$
${\displaystyle \left|x-4\right|>6}$
$x-4<-6orx-4>6$
$x<-6+4orx>6+4$
$x<-2orx>10$
Step 4
Therefore, the function is negative on the intervals$x<-2$ and $x>10.$
That is, the function is negative on ${\displaystyle (-\mathrm{\infty },-2)\text{and}(10,\mathrm{\infty })}$.
Thus, the other interval in which the function negative is ${\displaystyle (10,\mathrm{\infty })}$.