TMUA practice: Graph Transformations, problem 4

Build the target $\;y = 2(x - 1)^{2} + 3\;$ from $\;y = x^{2}\;$ by Maron’s rules I, II and IV:

• $(x - 1)^{2}$: replace $x$ by $x - 1$. The vertex moves from $(0,\, 0)$ to $(1,\, 0)$. (Rule I, horizontal translation.)
• $\;2(x - 1)^{2}\;$: multiply the output by $2$. Every $y$-value is doubled; in particular $0 \to 0$, so the vertex stays at $(1,\, 0)$. This is a vertical stretch by factor $2$ (rule IV).
• $\;2(x - 1)^{2} + 3\;$: add $3$ to every output. The vertex moves from $(1,\, 0)$ to $(1,\, 3)$. (Rule II, vertical translation.)

So the new graph has vertex $(1,\, 3)$ and a vertical stretch by factor $2$ relative to $y = x^{2}$. This is option A.

Why the other options fail.

• B has the sign of the horizontal translation wrong: $(x - 1)^{2}$ shifts right, putting the vertex at $x = +1$, not $x = -1$.
• C has the sign of the vertical translation wrong: $+3$ outside shifts up, so the $y$-coordinate of the vertex is $+3$, not $-3$.
• D has confused a vertical stretch with a horizontal compression. Multiplying the output by $2$ stretches vertically; the equation for a horizontal compression by $2$ would be $\;y = (2(x - 1))^{2} + 3 = 4(x - 1)^{2} + 3\;$, a different graph (vertical stretch by $4$, not $2$).
• E has used the coefficient $2$ in front as the $x$-coordinate of the vertex. But the $2$ is outside the bracket; it scales output, not input. The horizontal shift is controlled by $-1$ inside the bracket, putting the vertex at $x = 1$, not $x = 2$.

The lesson: in $\;y = a(x - h)^{2} + k\;$, the vertex is at $(h,\, k)$ and the factor $a$ outside is a vertical stretch (squash if $|a| < 1$, reflection in the $x$-axis if $a < 0$).