Turning Points by Completing the Square

Write the quadratic as a square plus a constant.

1. x^2+4x+9=(x+2)^2+5.
2. The square is smallest when x+2=0, so x=-2.
3. Then y=5, giving the turning point (-2,5).

Complete the square and find where the squared bracket is zero.

1. Complete the square: y=(x-4)^2+4.
2. The square is never negative, and it equals zero when x=4.
3. The minimum point is (4,4).

Set the squared bracket equal to zero before reading the other coordinate.

1. The square is zero when x-3=0, so x=3.
2. At this value of x, y=-7.
3. The turning point is (3,-7).
4. The tempting (-3,-7) has the wrong sign: (x-3)^2 is zero when x=+3, not -3.

Take out the negative sign before completing the square.

1. Write y=-(x^2-6x)-4.
2. Complete the square inside the bracket: y=-(x-3)^2+5.
3. The negative square is at most zero, with equality when x=3.
4. The maximum point is (3,5).

The smallest possible value of a squared bracket is zero.

1. Complete the square: x^2+10x+31=(x+5)^2+6.
2. The square is never negative and equals zero at x=-5.
3. Therefore the minimum value is 6.

Use the completed square to locate the turning point and decide which way the curve opens.

1. Complete the square: y=(x+1)^2-4.
2. The curve opens upwards and has a minimum at (-1,-4), left of the y-axis and below the x-axis.
3. It crosses the y-axis below the origin because y=-3 at x=0.
4. Sketch C has all these features.
5. A has its minimum to the right of the y-axis, B opens downwards, and D has its minimum above the x-axis.

A negative multiple of a square is at most zero.

1. Take out -2: y=-2(x^2-6x)+1.
2. Complete the square: y=-2(x-3)^2+19.
3. The term -2(x-3)^2 is at most zero.
4. So the maximum value of y is 19, reached when x=3.

The line of symmetry passes vertically through the turning point.

1. Complete the square: y=(x-\frac{3}{2})^2+\frac{11}{4}.
2. Equal distances on either side of x=\frac{3}{2} give equal values of the square.

3. Therefore the line of symmetry is x=\frac{3}{2}.

Factor out 2 from the terms containing x, then complete the square.

1. Write y=2(x^2+2x)-3.
2. Complete the square: y=2(x+1)^2-5.
3. The square is zero at x=-1, giving y=-5.
4. The turning point is (-1,-5).

Complete the square and tell apart the x-coordinate of the minimum point and the minimum value of y.

1. Write y=3(x^2-4x)+17.
2. Completing the square gives y=3(x-2)^2+5.
3. The squared term is never negative, so y has minimum 5 at x=2.
4. The statement "minimum value 2, reached when x=5" swaps the two numbers: 2 is where the minimum happens (x), and 5 is the minimum value (y).

Express the minimum value in terms of the unknown constant.

1. Complete the square: x^2+6x+c=(x+3)^2+c-9.
2. The minimum value is therefore c-9.
3. Set c-9=10 to obtain c=19.

A horizontal line meets the curve at only one point when it passes through the turning point.

1. Complete the square: y=2(x-2)^2+3.
2. The curve has a minimum value of 3.
3. A horizontal line meets this U-shaped curve at exactly one point when it passes through the minimum.

4. Therefore k=3.

The shift inside the square may be a fraction.

1. Factor out 4 from the first two terms: y=4(x^2-3x)+7.
2. Complete the square: y=4(x-\frac{3}{2})^2-2.
3. The square is zero at x=\frac{3}{2}.
4. The turning point is (\frac{3}{2},-2).

Complete the square and check that the best setting lies in the allowed range.

Part (a)

1. Complete the square: C=(s-7)^2+11.
2. The square is zero at s=7, which is an allowed setting.
3. The lowest cost is therefore reached at setting 7.

Part (b)

1. At s=7, the squared bracket is zero.
2. So the lowest cost is 11 pounds per hour.

Find the first turning point, then build the second curve around that same point.

Part (a)

1. The first curve is y=(x-5)^2-16, with turning point (5,-16).
2. The second curve must therefore be y=2(x-5)^2-16.

3. Expanding its linear term gives b=-20.

Part (b)

1. The second curve is y=2(x-5)^2-16.
2. Its constant term is 2\times25-16=34.
3. Therefore c=34.

Take out the negative leading coefficient and allow a fractional shift.

1. Reorder and take -2 out of the x terms: y=-2(x^2-\frac{5}{2}x)+3.
2. Half of -\frac{5}{2} is -\frac{5}{4}, so x^2-\frac{5}{2}x=(x-\frac{5}{4})^2-\frac{25}{16}.
3. Multiply the correction by -2: -2\times(-\frac{25}{16})=\frac{25}{8}, so y=-2(x-\frac{5}{4})^2+\frac{25}{8}+3.
4. \frac{25}{8}+3=\frac{49}{8}, so y=-2(x-\frac{5}{4})^2+\frac{49}{8}.
5. The squared term is zero when x=\frac{5}{4}, so the turning point (a maximum) is (\frac{5}{4},\frac{49}{8}).

Use the line of symmetry to write the squared bracket before finding the remaining constant.

1. The coefficient of x^2 is 1 and the symmetry line is x=-3, so write y=(x+3)^2+k.
2. Substitute the point (-1,8): 8=(-1+3)^2+k.
3. Thus k=4.
4. The turning point is (-3,4).

Complete the square to find the highest point, then use the same form at the line.

Part (a)

1. Write y=-(x^2-4x)+8.
2. Complete the square: y=-(x-2)^2+12.
3. The negative square is zero at x=2, giving the turning point (2,12).

Part (b)

1. Set the completed square form equal to 5: -(x-2)^2+12=5.
2. Rearrange to (x-2)^2=7.
3. The two x-coordinates are 2-\sqrt{7} and 2+\sqrt{7}.

Half of 4p is 2p: complete the square in terms of p.

1. Half of 4p is 2p, so x^2+4px=(x+2p)^2-4p^2.
2. So y=(x+2p)^2+3-4p^2.
3. The squared term is zero when x=-2p, and then y=3-4p^2.
4. The x^2 term is positive, so the curve opens upwards and the turning point is a minimum at (-2p,3-4p^2).
5. (2p,\ldots) has the wrong sign, since (x+2p)^2 is zero at x=-2p; and 3+4p^2 forgets that the 4p^2 added inside the square must be taken away.

Use the total fencing to express the area as a quadratic in one dimension.

1. The fencing is two lengths and three widths (two ends and the dividing fence): 2y+3x=36.
2. So y=18-\frac{3}{2}x, and the area is A=xy=18x-\frac{3}{2}x^2.
3. Take out -\frac{3}{2}: A=-\frac{3}{2}(x^2-12x).
4. Complete the square: x^2-12x=(x-6)^2-36, so A=-\frac{3}{2}(x-6)^2+54, because -\frac{3}{2}\times(-36)=54.
5. The term -\frac{3}{2}(x-6)^2 is never positive, so the greatest area is 54 m^2, when x=6 and y=9.