Successive transformations of a parabola

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Consider the parabola $y=x^2-6x+5$. Write the parabola transformed by each of the following (each applied to the original):

• (a) translation by $\vec{v}=(2;-3)$;
• (b) reflection in the $x$-axis;
• (c) dilation by factor $3$ along $y$;
• (d) rotation by $30^\circ$ counterclockwise about the origin.

Solution

Base: $y=(x-3)^2-4$, vertex $(3;-4)$.

(a) Substituting $x\to x-2,\ y\to y+3$: $y=x^2-10x+18$ (vertex $(5;-7)$).

(b) $y\to -y$: $y=-x^2+6x-5$.

(c) $y\to \dfrac{y}{3}$: $y=3x^2-18x+15$.

(d) With $\cos30^\circ=\tfrac{\sqrt3}{2}$, $\sin30^\circ=\tfrac12$, substitute $x\to \tfrac{\sqrt3}{2}x+\tfrac12 y$, $y\to -\tfrac12 x+\tfrac{\sqrt3}{2}y$: $-\tfrac12 x+\tfrac{\sqrt3}{2}y=\left(\tfrac{\sqrt3}{2}x+\tfrac12 y\right)^2-6\left(\tfrac{\sqrt3}{2}x+\tfrac12 y\right)+5.$

$\boxed{\text{(a) } y=x^2-10x+18;\ \text{(b) } y=-x^2+6x-5;\ \text{(c) } y=3x^2-18x+15}$