Expand (1/3)(x - 2)^2 - 3 into standard form.

Step-by-step solution

1. 1

   Square the bracket before multiplying by the fraction. Order of operations puts the exponent ahead of the multiplication:

   $(x-2)^2=(x-2)(x-2)=x^2-4x+4$

   Multiplying $\tfrac13$ into the bracket before squaring would be wrong — it would give $\left(\tfrac13 x-\tfrac23\right)^2$, a different expression.

2. 2

   Distribute the $\tfrac13$ across all three terms.

   $\frac{1}{3}\left(x^2-4x+4\right)=\frac{1}{3}x^2-\frac{4}{3}x+\frac{4}{3}$

3. 3

   Subtract the constant $3$ using a common denominator. Write $3=\tfrac93$:

   $\frac{4}{3}-3=\frac{4}{3}-\frac{9}{3}=-\frac{5}{3}$

4. 4

   Write the standard form.

   $\frac{1}{3}x^2-\frac{4}{3}x-\frac{5}{3}$

   The vertex form we started from tells us the vertex is at $(2,-3)$ and the parabola opens upward but is flattened by the factor $\tfrac13$.

5. 5

   Check at two values. At $x=2$: vertex form gives $0-3=-3$; standard form gives $\tfrac43-\tfrac83-\tfrac53=-3$ $\checkmark$. At $x=5$: vertex form gives $\tfrac13(9)-3=0$; standard form gives $\tfrac{25}{3}-\tfrac{20}{3}-\tfrac53=0$ $\checkmark$.