A11 - Turning points by completing the square (Higher)

A11 - Turning points by completing the square (Higher)

Completing the square rewrites a quadratic so that its turning point becomes visible. You will connect a squared bracket to a minimum or maximum, create that form from an expanded expression, and handle coefficients other than one. The result is exact, even when a plotted graph would only give an estimate.

Reading a turning point from a square (Higher)

What is the lowest possible value of y=x2+6x+5y=x^2+6x+5? Trying x=0x=0 gives 55, but negative inputs can take the graph lower. We want the exact lowest point without a long table.

You already know

An upward-opening quadratic has a minimum and a downward-opening one has a maximum. A square of a real number is non-negative. Expanding (x+p)2(x+p)^2 gives x2+2px+p2x^2+2px+p^2.

Start with y=(x−3)2−4y=(x-3)^2-4. Whatever xx is, the square is at least zero. Its smallest possible value is zero, reached when x−3=0x-3=0, so the smallest height is −4-4. You can locate a turn without differentiating, drawing, or finding any roots.

[DIAGRAM: asset_name: Turning points revealed by squared brackets; asset_slug: aqa8300_a22_square_form; caption: The bracket vanishes at the turning point, and the remaining constant gives its height; description: Two equal-scale coordinate panels showing y equals x minus 3 squared minus 4 and y equals x plus 3 squared minus 4. Minimum points 3 comma minus 4 and minus 3 comma minus 4, symmetry lines x equals 3 and x equals minus 3, and paired equal-height points one unit either side are labelled.]
The bracket vanishes at the turning point, and the remaining constant gives its height

Completed-square form

y=a(x−h)2+k,a≠0y=a(x-h)^2+k,\qquad a\ne0

The turning point is (h,k)(h,k) and the symmetry line is x=hx=h. If a>0a>0, kk is a minimum; if a<0a<0, kk is a maximum. These statements concern real xx.

Not on the formulae sheet: learn it.

The bracket is zero at x=hx=h. Moving equally far left and right gives squares of opposite numbers, which are equal. That explains both the turning point and the symmetry.

Worked example 1: a minus inside the bracket

Question. Write down the turning point of y=(x−3)2−4y=(x-3)^2-4. State whether it is a minimum or maximum. Non-calculator.

Plan. Make the squared bracket zero, then read the remaining height.

x−3=0smallest square is zerox=3add 3y=02−4=−4substitute the turning input\begin{aligned} x-3&=0 && \text{smallest square is zero}\\ x&=3 && \text{add }3\\ y&=0^2-4=-4 && \text{substitute the turning input} \end{aligned}

Why this step? The square cannot make yy smaller than −4-4, and it actually reaches zero at x=3x=3.

Answer. Turning point (3,−4)(3,-4), a minimum.

Check. At x=2x=2 and x=4x=4, y=−3y=-3, the same height on either side and above the minimum.

Change only the sign inside the bracket. The turning point moves to the opposite side of the vertical axis.

Worked example 2: a plus inside the bracket

Question. Write down the turning point of y=(x+3)2−4y=(x+3)^2-4. State whether it is a minimum or maximum. Non-calculator.

Plan. Solve x+3=0x+3=0, then evaluate the constant height.

x+3=0make the square zerox=−3subtract 3y=02−4=−4read the height\begin{aligned} x+3&=0 && \text{make the square zero}\\ x&=-3 && \text{subtract }3\\ y&=0^2-4=-4 && \text{read the height} \end{aligned}

Why this step? The turning input is the value that makes the bracket zero, rather than the number printed after xx.

Answer. Turning point (−3,−4)(-3,-4), a minimum.

Check. At x=−4x=-4 and x=−2x=-2, the square equals 11, so both heights are −3-3.

A leading minus changes what “smallest square” means for the output. For y=−(x+3)2−4y=-(x+3)^2-4, the squared term is still non-negative, but its negative is non-positive. The turning point remains (−3,−4)(-3,-4), now a maximum.

Watch out

Wrong: reading the turning point of y=(x+3)2−4y=(x+3)^2-4 as (3,−4)(3,-4). Right: solve x+3=0x+3=0 to get −3-3. The sign inside the bracket is reversed when you name the turning input.

Use the same zero-bracket decision with different values.

Creating the squared bracket (Higher)

The middle term tells you what must go inside the square. Expanding (x+3)2(x+3)^2 gives x2+6x+9x^2+6x+9. So x2+6x+5x^2+6x+5 is that square with 44 taken away. You must correct the constant, because rewriting an expression must preserve its value for every xx.

Completing the square with leading coefficient one

x2+bx+c=(x+b2)2+c−(b2)2x^2+bx+c=\left(x+\frac b2\right)^2+c-\left(\frac b2\right)^2

For real bb and cc, halve the coefficient of xx, square that half, and subtract the extra constant introduced.

Not on the formulae sheet: learn it.

The correction has a reason: the bracket contributes (b/2)2(b/2)^2 as well as x2+bxx^2+bx. Subtracting it again keeps the original expression unchanged. An odd middle coefficient simply gives a fractional half.

Worked example 3: an even middle coefficient

Question. Work out the turning point of y=x2+6x+5y=x^2+6x+5 by completing the square. Non-calculator.

Plan. Use half of 66 in the bracket, correct its constant, then make the square zero.

y=(x+3)2−9+5add and remove 32=(x+3)2−4combine constantsx+3=0⟹x=−3make the square zeroy=−4remaining constant\begin{aligned} y&=(x+3)^2-9+5 && \text{add and remove }3^2\\ &=(x+3)^2-4 && \text{combine constants}\\ x+3&=0\quad\Longrightarrow\quad x=-3 && \text{make the square zero}\\ y&=-4 && \text{remaining constant} \end{aligned}

Why this step? (x+3)2(x+3)^2 already contains 99; subtracting 99 prevents that extra amount changing the function.

Answer. Turning point (−3,−4)(-3,-4), a minimum.

Check. Expanding gives x2+6x+9−4=x2+6x+5x^2+6x+9-4=x^2+6x+5. At x=−3x=-3, the original gives 9−18+5=−49-18+5=-4.

Change only the middle coefficient from 66 to 55. The method is unchanged, but keep the half and its square as exact fractions.

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