LEARN · EXPLAIN · REVISE

Complete the square. Locate the vertex and find the minimum value of f over all real x.

Read the idea, work independently, then explain what changed.

TOPIC 01

Quadratic equations and functions

Expanding 2(x − 3)² − 5 recovers the original function. The factor 2 stretches the graph vertically without changing the x-coordinate of the vertex.

PREDICT → MOVE → EXPLAIN

What does each coefficient change?

Before moving a slider, predict the opening, vertex and intercepts. Then compare the graph with your prediction.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Worked example

Complete the square. Locate the vertex and find the minimum value of f over all real x.

f(x)=2x^2-12x+13

Working and explanation

BUILD THE REASONING

Hint 1

Factor 2 out of the two terms containing x before completing the square.

Hint 2

Half of −6 is −3. Add and subtract 9 inside the bracket.

Worked solution

Isolate the quadratic and linear terms.
$f(x)=2(x^2-6x)+13$
Make a perfect square while keeping the expression equal.
$f(x)=2[(x-3)^2-9]+13=2(x-3)^2-5$
The square is smallest at x = 3. The parabola opens upwards.
$(h,k)=(3,-5),\qquad \min f=-5$

Vertex (3, −5); axis x = 3; minimum −5 at x = 3.

Checks and common pitfalls: Expanding 2(x − 3)² − 5 recovers the original function. The factor 2 stretches the graph vertically without changing the x-coordinate of the vertex.

Think first. Reveal a hint when the class is ready.