TMUA practice: Graph Transformations, problem 3

Compare the two factorisations.

• $\;f(x) = (x + 1)(x - 2)(x - 5)\;$ has zeros at $x = -1,\, 2,\, 5$.
• $\;g(x) = (x + 4)(x + 1)(x - 2)\;$ has zeros at $x = -4,\, -1,\, 2$.

Each zero of $g$ is exactly $3$ less than the corresponding zero of $f$:

$-1 \to -4, \quad 2 \to -1, \quad 5 \to 2 \quad (\text{each subtracts } 3).$

This is consistent with a translation $3$ units to the left, which corresponds to replacing $x$ by $x + 3$ in $f$.

Verify by substitution:

The identity holds for every $x$, so $g(x) = f(x + 3)$, and the graph of $g$ is the graph of $f$ translated $3$ units to the left. This is option C.

Why the other options fail.

• A has the direction wrong. $f(x - 3)$ shifts right by $3$, sending the zeros to $-1 + 3 = 2$, $\; 2 + 3 = 5$, $\; 5 + 3 = 8$, which would give roots $\{2,\, 5,\, 8\}$, not $\{-4,\, -1,\, 2\}$.
• B, reflection in $y$-axis, would send the zeros to $\{1,\, -2,\, -5\}$, again not the zeros of $g$.
• D, $f(x) + 3$, does not change the $x$-coordinates of the zeros; for instance $f(-1) + 3 = 0 + 3 = 3 \neq 0$, so $-1$ is not a zero of $f(x) + 3$. But $-1$ is a zero of $g$.
• E, $-f(x)$, has the same zeros as $f$, namely $\{-1,\, 2,\, 5\}$, not $\{-4,\, -1,\, 2\}$.

The lesson: a horizontal translation by $h$ shifts every zero by $h$ (and preserves multiplicities). Comparing two sets of zeros that all differ by the same constant is a clean tell for a horizontal translation.