We want the distance between $x$ and 5 to be less than or equal to 4. The distance can be represented using the absolute value, giving the expression

$$|x−5| ≤ 4$$

Since we want the size of the difference between the actual percentage, $p$, and the reported percentage to be less than 3%,

$$|p−78| \leq 3$$

First, we must recognize the sharp V-shape of the graph as an absolute value function, $g(x)=|x|$. We’ll need to find $a, $b, $c$, and $d$, so we can write our function in the form

$$f(x)=ag(bx+c)+d,$$

that is,

$$f(x)=a|bx+c|+d.$$

The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 and down 2 from the basic toolkit function. This tells us $c= -3$ and $d=2$.

We might also notice that the graph appears stretched, since the linear portions have slopes of 2 and -2. From this information we could write the write the equation in two ways:

$$f(x)=2|x−3| -2,$$ treating the stretch as a vertical stretch

$$f(x)=|2(x−3)| -2,$$ treating the stretch as a horizontal compression

Note that these equations are algebraically equivalent – the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch/compression.

If you had not been able to determine the stretch based on the slopes of the lines, you can solve for the stretch factor by putting in a known pair of values for $x$ and $f(x)$.

$$f(x)=a|x−3| -2$$ Now substituting in the point (1, 2)

$$2=a|1−3| -2 $$ $$4=2a$$ $$a=2$$

Taking a transformations perspective, we first notice the absolute value bars. That tells us our parent function is $f(x)=|x|$. If we match up our transformations to the form $g(x)=af(bx+c)+d=a|bx+c|+d$, we have $a= -1$, $c=1$, and $d=2$. We’ll apply $c$ first to create a shift left 1 unit, then apply $a$ to create a vertical reflection. We’ll end with $d, a shift up 2. This process creates the following table and graph.

$$\begin{array}{|c|c|c|c|c|} \hline X=x-1 & \leftarrow x & f(x) \rightarrow & y=-1\cdot f(x)\rightarrow & g(X)=y+2\\ \hline -3 & -2 & 2 & -2 & 0\\ \hline -2 & -1 & 1 & -1 & 1\\ \hline -1 & 0 & 0 & 0 & 2\\ \hline 0 & 1 & 1 & -1 & 1\\ \hline 1 & 2 & 2 & -2 & 0\\ \hline \end{array}$$

This function has domain $(−\infty,\infty)$ and range $( -\infty, 2]$.

The $x$-intercepts are $(−3,0)$ and $(1,0)$. The $y$-intercept is $(0,1$).

The end behavior is as $x\to\infty, g(x)\to-\infty$ and as $x\to-\infty, g(x)\to-\infty$.

The graph is increasing on $(−\infty, −1)$ and decreasing on $(−1, \infty)$.

The function has an absolute maximum value of 2 at $x= -1$.

The horizontal intercepts will occur when $f(x)=0$. Solving,

\[ \begin{array}{rlc} 0 &=|4x+1|-7 & \text{ Isolate the absolute value} \\ 7 &=|4x+1| & \text{ Now break this into two equations:} \\ \end{array} \] \[ \begin{array}{rl} 7 &=4x+1 \quad \text{or} \quad -7=4x+1 \\ 6 &=4x \quad \text{or} \quad -8=4x \\ x &=\frac{6}{4} \quad \text{or} \quad x=\frac{-8}{4} \\ x &=\frac{3}{2} \quad \text{or} \quad x=-2 \end{array} \]

The graph has two horizontal intercepts, at $x=\frac{3}{2}$ and $x =  -2$.

Isolating the absolute value on one side the equation,

$$1=4|x−2|+2$$

$$ -1=4|x−2|$$

$$-\frac{1}{4}=|x-2|$$

At this point, we notice that this equation has no solutions – the absolute value always returns a positive value, so it is impossible for the absolute value to equal a negative value.

With both approaches, we will need to know first where the corresponding equality is true. In this case, we first will find where $|x−5|=4$. We do this because the absolute value is a nice friendly function with no breaks, so the only way the function values can switch from being less than 4 to being greater than 4 is by passing through where the values equal 4. Solve $|x−5|=4$,

$$\begin{array}{c} x-5=4 \\ x=9 \end{array}\ \ \text{or}\ \ \begin{array}{c} x-5=- 4 \\ x=1 \end{array}$$

To use a graph, we can sketch the function $f(x)=|x−5|$. To help us see where the outputs are 4, the line $g(x)=4$ could also be sketched.

On the graph, we can see that indeed the output values of the absolute value are equal to 4 at $x=1$ and $x=9$. Based on the shape of the graph, we can determine the absolute value is less than or equal to 4 between these two points, when $1 \leq x \leq 9$. In interval notation, this would be the interval $[1,9]$.

As an alternative to graphing, after determining that the absolute value is equal to 4 at $x=1$ and $x=9$, we know the graph can only change from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals: $x\lt 1$, $1\lt x\lt 9$, and $x\gt 9$. To determine when the function is less than 4, we could pick a value in each interval and see if the output is less than or greater than 4.

$$\begin{array}{cccc} \text{Interval} & \text{Test }x & f(x) & \text{less than or greater than 4?}\\ \hline x\lt 1 & 0 & |0-5|=5 & \text{greater}\\ 1 \lt x \lt 9 & 6 & |6-5|=1 & \text{less}\\ x \gt 9 & 11 & |11-5|=6 & \text{greater}\\ \end{array}$$

Since $1 \leq x \leq 9$ is the only interval in which the output at the test value is less than 4, we can conclude the solution to $|x−5| \leq 4$ is $1 \leq x \leq 9$.

We are trying to determine where $f(x) \lt 0$, which is when $- \frac{1}{2}|4x-5|+3 \lt 0$. We begin by isolating the absolute value:

$$- \frac{1}{2}|4x-5| \lt-3$$ when we multiply both sides by -2, it reverses the inequality

$$|4x−5| \gt 6$$ 

Next we solve for the equality $|4x−5|=6$

$$\begin{array}{c} 4x-5=6 \\ 4x=11 \\ x=\frac{11}{4} \end{array}\ \ \ \text{ or }\ \ \ \ \begin{array}{c} 4x-5=-6 \\ 4x=-1 \\ x=-\frac{1}{4} \end{array}$$

We can now either pick test values or sketch a graph of the function to determine on which intervals the original function value are negative. Notice that it is not even really important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at $x=\frac{-1}{4}$ and $x=\frac{11}{4}$, and that the graph has been reflected vertically due to the negative $a$-value.

From the graph of the function, we can see the function values are negative to the left of the first horizontal intercept at $x=\frac{-1}{4}$, and negative to the right of the second intercept at $x = \frac{11}{4}$. This gives us the solution to the inequality:

$$x \lt \frac{-1}{4}\quad \text{ or }\quad x \gt \frac{11}{4}$$

In interval notation, this would be $$\left( -\infty,\frac{-1}{4} \right) \cup \left( \frac{11}{4},\infty \right)$$.