Question

An absolute value function with a vertex of (4,14) is negative on the interval \(\displaystyle{\left(−\infty,−{2}\right)}\).
On what other interval is the function negative?

Answer 1

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 \(\displaystyle{\left({10},\infty\right)}\).