Alyce Wilkinson

2021-03-11

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

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

Step 4
Therefore, the function is negative on the intervals$x<-2$ and $x>10.$
That is, the function is negative on $\left(-\mathrm{\infty },-2\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(10,\mathrm{\infty }\right)$.
Thus, the other interval in which the function negative is $\left(10,\mathrm{\infty }\right)$.

Jeffrey Jordon