TMUA practice: Graph Transformations, problem 1

Let $\;g(x) = f(2x) + 1$. We are told that $\;f(-2) = 5\;$ and $\;f(1) = -3$. To use these, set $2x = -2$ and $2x = 1$ in turn.

• $2x = -2$ gives $x = -1$, and $\;g(-1) = f(-2) + 1 = 5 + 1 = 6$. So $(-1,\, 6)$ is on the graph of $g$.
• $2x = 1$ gives $x = \tfrac{1}{2}$, and $\;g\!\left(\tfrac{1}{2}\right) = f(1) + 1 = -3 + 1 = -2$. So $\left(\tfrac{1}{2},\, -2\right)$ is on the graph of $g$.

The two points are $(-1,\, 6)$ and $\left(\tfrac{1}{2},\, -2\right)$. This is option C.

Geometrically: $y = f(2x) + 1$ compresses the graph of $f$ horizontally by factor $2$ (the $x$-coordinates halve) and shifts it up by $1$. The point $(-2,\, 5)$ moves to $(-1,\, 6)$; the point $(1,\, -3)$ moves to $(\tfrac{1}{2},\, -2)$.

Why the other options fail.

• A has the $x$-coordinates doubled (corresponding to $f(x/2)$, a horizontal stretch) instead of halved.
• B has the $x$-coordinates halved but has not added $1$ to the $y$-coordinates.
• D has not changed the $x$-coordinates (corresponding to $f(x) + 1$, no horizontal scaling).
• E has the first $x$-coordinate halved correctly but has flipped the sign of the second: $\;\tfrac{1}{2}\;$ became $\;-\tfrac{1}{2}$. The transformation $y = f(2x)$ involves no sign change.

The lesson: $y = f(2x) + 1$ acts on coordinates as $(x_{0},\, y_{0}) \mapsto (\tfrac{x_{0}}{2},\, y_{0} + 1)$. Half the $x$-coordinate (because the input doubles inside $f$) and add $1$ to the $y$-coordinate.