TMUA practice: Graph Transformations, problem 2

Let $\;g(x) = f(x - 2) + 3$. The graph of $g$ is the graph of $f$ translated by the vector $(2,\, 3)$ — $2$ units right (because of $x - 2$ inside) and $3$ units up (because of $+3$ outside).

Translation preserves all distances and so preserves every axis of symmetry. The graph of $f$ is symmetric about the $y$-axis, which is the vertical line $x = 0$. After translating $2$ units right, this axis becomes the vertical line $x = 0 + 2 = 2$. (The upward translation by $3$ moves points along the symmetry axis but does not change the axis itself, because a vertical line is unchanged by a vertical translation.) So the new graph is symmetric about $x = 2$. This is option B.

Direct algebraic check. For any real $h$:

$g(2 + h) \;=\; f((2 + h) - 2) + 3 \;=\; f(h) + 3,$

$g(2 - h) \;=\; f((2 - h) - 2) + 3 \;=\; f(-h) + 3.$

Since $f$ is even, $f(-h) = f(h)$, so $g(2 + h) = g(2 - h)$ for every $h$. That is exactly the condition for symmetry about $x = 2$.

Why the other options fail.

• A is wrong: the horizontal translation by $2$ has moved the axis of symmetry away from the $y$-axis. The graph of $g$ does not satisfy $g(-x) = g(x)$ in general.
• C has the wrong sign: $f(x - 2)$ shifts right by $2$, so the new axis sits at $x = 2$, not $x = -2$.
• D is a horizontal line; the graph of a non-constant function cannot be symmetric about a horizontal line.
• E is wrong because the axis of symmetry $x = 0$ of $f$ has clearly moved to $x = 2$ under the translation; it has not vanished.

The lesson: a translation moves any axis of symmetry by the same vector. A vertical line moves only horizontally; a horizontal line moves only vertically. So the horizontal component of the translation governs where a vertical axis of symmetry ends up.