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 \(\displaystyle{f{{\left({x}\right)}}}={\left\lbrace\begin{array}{cc} {a}{\left({x}-{4}\right)}+{14}&{x}{>}{4}\\-{a}{\left({x}-{4}\right)}+{14}&{x}{<}{4}\end{array}\right.}\)</span>

Step 2

The function is negative on the interval \(\displaystyle{\left(−\infty,−{2}\right)}\). 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.

\(\displaystyle-\frac{{7}}{{3}}{\left|{{x}-{4}}\right|}+{14}{<}{0}\)</span>

\(\displaystyle-\frac{{7}}{{3}}{\left|{{x}-{4}}\right|}{<}-{14}\)</span>

\(\displaystyle-\frac{{7}}{{3}}{\left|{{x}-{4}}\right|}{>}{14}\)

\(\displaystyle{3}\cdot\frac{{7}}{{3}}{\left|{{x}-{4}}\right|}{>}{14}\cdot{3}\)

\(\displaystyle{7}{\left|{{x}-{4}}\right|}{>}{42}\)

\(\displaystyle{\left|{{x}-{4}}\right|}{>}{6}\)

x-4<-6 or x-4 > 6

x<-6+4 or x>6+4

x<-2 or x>10

Step 4

Therefore, the function is negative on the intervals x<−2 and x>10.

That is, the function is negative on \(\displaystyle{\left(−\infty,−{2}\right)}{\quad\text{and}\quad}{\left({10},\infty\right)}\).

Thus, the other interval in which the function negative is \(\displaystyle{\left({10},\infty\right)}\).

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 \(\displaystyle{f{{\left({x}\right)}}}={\left\lbrace\begin{array}{cc} {a}{\left({x}-{4}\right)}+{14}&{x}{>}{4}\\-{a}{\left({x}-{4}\right)}+{14}&{x}{<}{4}\end{array}\right.}\)</span>

Step 2

The function is negative on the interval \(\displaystyle{\left(−\infty,−{2}\right)}\). 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.

\(\displaystyle-\frac{{7}}{{3}}{\left|{{x}-{4}}\right|}+{14}{<}{0}\)</span>

\(\displaystyle-\frac{{7}}{{3}}{\left|{{x}-{4}}\right|}{<}-{14}\)</span>

\(\displaystyle-\frac{{7}}{{3}}{\left|{{x}-{4}}\right|}{>}{14}\)

\(\displaystyle{3}\cdot\frac{{7}}{{3}}{\left|{{x}-{4}}\right|}{>}{14}\cdot{3}\)

\(\displaystyle{7}{\left|{{x}-{4}}\right|}{>}{42}\)

\(\displaystyle{\left|{{x}-{4}}\right|}{>}{6}\)

x-4<-6 or x-4 > 6

x<-6+4 or x>6+4

x<-2 or x>10

Step 4

Therefore, the function is negative on the intervals x<−2 and x>10.

That is, the function is negative on \(\displaystyle{\left(−\infty,−{2}\right)}{\quad\text{and}\quad}{\left({10},\infty\right)}\).

Thus, the other interval in which the function negative is \(\displaystyle{\left({10},\infty\right)}\).