Vertex form and completing the square

1. Preserve equality.

   $$f=(x-3)^2-4$$

2. Bound the square and identify equality.

   $$(x-3)^2\ge0,\quad x=3\text{ at equality}$$

Minimum −4 at x=3.

Checks and common pitfalls: The value and its attaining input are different parts of the answer.

1. Use vertex form.

   $$y=a(x-2)^2-3$$

2. Determine the scale from the extra point.

   $$5=4a-3\Rightarrow a=2$$

y=2(x−2)²−3.

Checks and common pitfalls: The extra point has a different x-coordinate, so it determines a uniquely.

1. Represent positive side lengths.

   $$x,10-x,\quad0<x<10$$

2. The square gives a bound with a feasible equality case.

   $$A=25-(x-5)^2\le25,\quad x=5$$

25 m², attained by a 5 m square.

Checks and common pitfalls: Both the model and equality case must respect positive side lengths.

1. Locate the vanishing square.

   $$x+1=0$$

2. Solve and check the height.

   $$x=-1,\quad y=-5$$

−1.

Checks and common pitfalls: x+1 means x−(−1).

1. Keep the external coefficient.

   $$y=2[(x+2)^2-4]+3$$

2. Collect the full compensation.

   $$y=2(x+2)^2-5$$

Vertex (−2,−5).

Checks and common pitfalls: Subtracting only 4 rather than 8 would change the function.

1. Check symmetric inputs.

   $$(1-4)^2=(7-4)^2$$

2. Use equality of the outputs.

   $$f(7)=f(1)=2$$

2.

Checks and common pitfalls: Symmetry avoids reconstructing unnecessary coefficients.

1. Bound the variable term.

   $$-3(x-2)^2\le0$$

2. Check equality is attained.

   $$f\le12,\quad f(2)=12$$

12.

Checks and common pitfalls: The negative coefficient makes this a maximum.

1. Compare distances to the axis.

   $$|1-3|=2>|4-3|=1$$

2. Squaring preserves their nonnegative order.

   $$f(1)=0>f(4)=-3$$

f(1)>f(4).

Checks and common pitfalls: Subtracting a common constant preserves the comparison.

1. Write the resulting vertex form.

   $$f=ax^2+2$$

2. Use and check noncentral points.

   $$3=a+2\Rightarrow a=1,\quad f(2)=6$$

y=x²+2.

Checks and common pitfalls: The stated quadratic assumption matters; five values do not determine an arbitrary function.

1. Apply horizontal and vertical changes separately.

   $$y=(x-2)^2+3$$

2. Locate the zero square.

   $$(h,k)=(2,3)$$

y=(x−2)²+3; vertex (2,3).

Checks and common pitfalls: A right shift subtracts inside the input.

1. Identify the parameter-dependent minimum.

   $$f=(x-k)^2+9-k^2$$

2. Solve and apply k>0.

   $$9-k^2=5\Rightarrow k=2$$

2.

Checks and common pitfalls: The sign restriction removes the second algebraic candidate.

1. All candidates have this factor form and axis.

   $$f=a(x-1)(x-5),\quad h=3$$

2. Match the minimum; the unique a is positive.

   $$-4a=-8\Rightarrow f=2(x-3)^2-8$$

Uniquely f=2(x−3)²−8.

Checks and common pitfalls: This joins intercept and extremum information instead of repeating a single representation.

1. Expose the real bound.

   $$C=(n-4.5)^2+9.75$$

2. Use the smallest allowed distance.

   $$C(4)=C(5)=10$$

4 or 5 packs; minimum cost 10.

Checks and common pitfalls: Every other allowed integer is farther from 4.5, proving completeness.