Absolute Value Equations and Graphs: Free Response

Passing means the weight $w$ sits no farther than $25$ grams from the target $900$ grams, on either side. The distance between $w$ and $900$ is $\lvert w - 900 \rvert$, so requiring that distance to be at most $25$ grams gives

$\lvert w - 900 \rvert \le 25.$

The lightest passing loaf sits $25$ grams below the target, and the heaviest sits $25$ grams above it. Subtracting and adding $25$ from the target $900$ gives

$900 - 25 = 875, \qquad 900 + 25 = 925,$

so a passing loaf weighs at least $875$ grams and at most $925$ grams.

Compute the actual distance between the loaf's weight and the target:

$\lvert 918 - 900 \rvert = \lvert 18 \rvert = 18.$

Since $18 \le 25$, the loaf's weight sits within the allowed tolerance, so it passes quality control.

Since $x = -17$ is negative, the piecewise rule uses the second branch: $\lvert x \rvert = -x = -(-17) = 17$. Since $y = 9$ is already zero or positive, the first branch applies directly: $\lvert y \rvert = y = 9$.

$\lvert -17 \rvert = 17, \qquad \lvert 9 \rvert = 9.$

The symbol $-x$ means 'the opposite of $x$,' not 'a negative number.' When $x$ itself is already negative, its opposite is positive, because flipping a negative sign lands on the positive side of zero:

$x = -17 \;\Longrightarrow\; -x = -(-17) = 17.$

The two minus signs, the one already carried by $x$'s own value and the one written in front of it in the rule, cancel, leaving a positive result.

Comparing the two results from part A,

$\lvert -17 \rvert = 17 > 9 = \lvert 9 \rvert,$

so $x = -17$ sits farther from zero than $y = 9$. This decision rests only on the SIZES of the two absolute values; it has nothing to do with which original number carried a negative sign or which one looked bigger as a signed value.

Match term by term to $y = a\lvert x - h \rvert + k$. The coefficient in front of the bars is $a = -3$. Inside the bars, $x - 4$ matches $x - h$ directly, so $h = 4$. The constant added outside is $k = 7$.

The vertex sits where the bars equal zero, at $x = h$, giving output $y = a\cdot 0 + k = k$:

$(h, k) = (4, 7).$

An absolute value accepts any real input, since every number has some distance from zero, and none of the constants $a$, $h$, or $k$ restricts which inputs are allowed: they only relocate, stretch, or flip the OUTPUT. So the domain stays all real numbers no matter what the equation's constants are.

The range depends on which way the V opens. Here $a = -3 < 0$, so the graph opens downward from the vertex, meaning the vertex is the highest point and every other output is smaller:

$y \le k = 7.$

Substitute $x=6$ into the equation:

$y = -3\lvert 6 - 4 \rvert + 7 = -3(2) + 7 = 1.$

This height, $1$, is below the vertex height of $7$. Moving away from the vertex in either direction makes the bars a positive number, and multiplying a positive number by the negative $a=-3$ only ever subtracts from $k$. So no point away from the vertex can reach or exceed the vertex's own height of $7$.

The bars split into two cases, one for each sign of the quantity inside:

$x - 6 = 3x - 2 \quad \text{or} \quad x - 6 = -(3x - 2).$

The first case gives $x - 3x = -2 + 6$, so $-2x = 4$ and $x = -2$. The second case gives $x - 6 = -3x + 2$, so $4x = 8$ and $x = 2$.

Substitute each candidate back into the ORIGINAL equation, $\lvert x - 6 \rvert = 3x - 2$, not the split linear equation that produced it.

For $x=2$, both sides agree:

$\lvert 2-6 \rvert = 4 = 3(2)-2,$

so $x=2$ checks. For $x=-2$, the two sides do not agree:

$\lvert -2-6 \rvert = 8, \qquad 3(-2)-2 = -8,$

so $x=-2$ fails.

The split

$X = R \quad \text{or} \quad X = -R$

solves each branch as an ordinary linear equation, with no reference at all to the SIGN of $R$ at the candidate that results.

But $\lvert X \rvert$ itself can never be negative, so any candidate that makes $R$ come out negative cannot possibly be genuine, however cleanly it solved its linear equation. The only way to catch this is to substitute each candidate back into the ORIGINAL equation, $\lvert X \rvert = R$, and confirm both sides actually agree, exactly as done in part B.

Only the number $0$ itself is zero units from zero, so $\lvert X \rvert = 0$ forces

$X = 0,$

a single solution. The claim's own pattern, $X=c$ and $X=-c$, would give $X=0$ and $X=-0$, which are the SAME point, not two distinct ones.

An absolute value is a distance, so $\lvert X \rvert \ge 0$ for every real $X$. Since $-6$ is negative, no value of $X$ can make $\lvert X \rvert$ equal $-6$:

$\lvert X \rvert \ge 0 > -6.$

So $\lvert X\rvert = -6$ has zero solutions, not two.

The two tests above already rule out the unconditional claim: $c=0$ gave one solution, and $c=-6$ gave none, not two either time. Sorting by the sign of $c$ accounts for every case:

$c>0: \text{ two solutions}, \qquad c=0: \text{ one solution}, \qquad c<0: \text{ no solution}.$

The corrected claim is not 'always two,' but 'two solutions exactly when $c>0$'; the sign of $c$ has to be checked before any splitting begins.