Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — completing the square and reading the minimum
Express x2−4x+3 in the form (x+p)2+q and state the minimum point.
- 1Halve the coefficient of x: half of −4 is −2.The sign is kept.
- 2x2−4x=(x−2)2−4(x−2)2 expands to x2−4x+4, so 4 must be taken back off.
- 3Bring the +3 along: (x−2)2−4+3
- 4=(x−2)2−1Check by expanding: x2−4x+4−1 ✓
- 5Minimum point (2,−1)The bracket is smallest — zero — when x=2, and then y=−1.
Example 2 — a negative quadratic
Express 5+4x−x2 in completed-square form and state the maximum point.
- 1Take the minus out of the x terms: −(x2−4x)+5Both signs inside the bracket change.
- 2x2−4x=(x−2)2−4Complete the square inside as usual.
- 3−[(x−2)2−4]+5
- 4=−(x−2)2+4+5=−(x−2)2+9The −4 inside becomes +4 outside.
- 5Maximum point (2,9)Check in the original: 5+8−4=9 ✓. Negative x2 means a maximum.
Example 3 — where a line cuts a curve
Find the points where y=x2 meets y=x+2.
- 1At an intersection both equations give the same y, so set them equal.
- 2x2=x+2
- 3x2−x−2=0Rearrange to equal zero before factorising.
- 4(x−2)(x+1)=0, so x=2 or x=−1Two numbers multiplying to −2 and adding to −1.
- 5x=2⇒y=4; x=−1⇒y=1Substitute into the simpler equation.
- 6The points are (2,4) and (−1,1).Check in y=x2: 22=4 ✓ and (−1)2=1 ✓