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# 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? # 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?

Question
Piecewise-Defined Functions asked 2021-03-11
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?

## Answers (1) 2021-03-12
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 $$\displaystyle{\left({10},\infty\right)}$$.

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