Completing the square maxima and minima - …

1 Completing the squaremaxima and minimamc-TY-completingsquare1-2009-1 Completing the square is an algebraic technique which has several applications. These includethe solution of quadratic equations . In this unit we use it tofind the maximum or minimumvalues of quadratic order to master the techniques explained here it is vital that you undertake plenty of practiceexercises so that all this becomes second nature. To help youto achieve this, the unit includesa substantial number of such reading this text, and/or viewing the video tutorial on this topic, you should be able to: complete the square for a quadratic expression find maximum or minimum values of a quadratic function by Completing the minimum value of a quadratic is meant by a complete, or exact, square ?

2 The square - when the coefficient ofx2is the square - when the coefficient ofx2is not mathcentre 20091. IntroductionCompleting the Squareis a technique which can be used to find maximum or minimum valuesof quadratic functions. We can also use this technique to change or simplify the form of algebraicexpressions. We can use it for solving quadratic this unit we will be using Completing the square to find maximum and minimum values ofquadratic The minimum value of a quadratic functionConsider the functiony=x2+ 5x 2 You may be aware from previous work that the graph of a quadratic function, where the coefficientofx2is positive as it is here, will take the form of one of the graphs shown in Figure 1.

3 We may be interested in finding the coordinates of the minimum pointThere will be a minimum, or lowest point on the graph and we maybe interested in findingthexandyvalues at this point. Is the minimum point below or above the horizontal axis ?This question can be answered using techniques in calculus,but here as an alternative we useCompleting the What is meant by a complete or exact square ?An expression of the form(x+a)2is called acompleteorexactsquare. Multiplying this out we obtain(x+a)2= (x+a)(x+a)=x2+ 2ax+a2In the same way, consider(x a)2 Multiplying out the brackets:(x a)2= (x a)(x a)=x2 2ax+a2 Both expressionsx2+ 2ax+a2andx2 2ax+a2are called complete squares because they canbe written as a single term squared, that is(x+a)2, or(x a) mathcentre 20094.

4 Completing the square - when the coefficient ofx2is 1We now return to the quadratic expressionx2+ 5x 2and we are going to try to write it in theform of a single term squared, that is a complete square , in this case(x+a) the two expressions:x2+ 2ax+a2andx2+ 5x 2 Clearly the coefficients ofx2in both expressions are the same - they match would like to match up the term2axwith the term5x. To do this note that2amust equal5, so thata= that(x+a)2=x2+ 2ax+a2 With this value ofa(x+52)2=x2+ 5x+(52)2 However, the right hand side is not the same as the original expressionx2+ 5x 2.

5 To make itthe same we need to subtract(52)2to remove this unwanted term:(x+52)2 (52)2=x2+ 5x+ (52)2 (52)2and then subtract 2 to insert the term we need:(x+52)2 (52)2 2 =x2+ 5x 2So we havex2+ 5x 2 =(x+52)2 (52)2 2 Combining the last two numbersx2+ 5x 2 =(x+52)2 334At this stage we have finished Completing the square for the quadratic expressionx2+ 5x expression is not a complete or exact square because in addition to the complete square (x+52)2there is the constant term written the expressionx2+5x 2in this form it is now straightforward to find its minimumvalue.

6 This is because the first part,(x+52)2being a square , is always positive unless it is equalto zero. Zero is its lowest possible value and so the lowest possible value ofx2+ 5x 2must be 334. This lowest value will occur when(x+52)2is zero, that is whenx= mathcentre 2009In conclusion, we have shown that the minimum value ofx2+ 5x 2is 334and this occurswhenx= we wish to find the minimum value of the quadratic functionf(x) =x2 6x the two expressions:x2 2ax+a2andx2 6x 12 Clearly the coefficients ofx2in both expressions are the same - they match would like to match up the term 2axwith the term 6x.

7 To do this note that 2amustbe 6, so thata= that(x a)2=x2 2ax+a2 With this value ofa(x 3)2=x2 6x+ 9 However the right hand side is not yet the same as our originalfunctionx2 6x 12. To makeit the same we subtract 9 from both sides:(x 3)2 9 =x2 6x+ 9 9and then subtract 12 from each side:(x 3)2 9 12 =x2 6x 12So we havex2 6x 12 = (x 3)2 9 12= (x 3)2 21We havecompleted the square . We have writtenx2 6x 12as a complete square (x 3)2together with an additional term know that becausef(x) =x2 6x 12has a positivex2term the graph will have a will occur when(x 3)2is zero.

8 The minimum value off(x)will be 21whenx= Complete the square for each of the following expressionsa)x2+ 6x+ 3b)x2 10x 6c)x2+ 20x+ 100d)x2 5x+ 2e)x2+x+ 1f)x2 x+ 12. Find the minimum values of the following expressionsa)x2 x 1b)x2+x 1c)x2+ 2x+ 1d)x2 8x+ 5e)x2+12x+12f)x2 45x+ mathcentre 20095. Example where the coefficient ofx2is not 1We now consider a more complicated example where the coefficient ofx2is not (x) = 2x2 6x+ first step is to take the 2 out as a common factor as follows:2x2 6x+ 1 = 2(x2 3x+12)We now complete the square as before with the bracketed the two expressions:x2 2ax+a2andx2 3x+12 Clearly the coefficients ofx2in both expressions are the same - they match would like to match up the term 2axwith the term 3x.

9 To do this note that2amust be3, so thata= that(x a)2=x2 2ax+a2 With this value ofa(x 32)2=x2 3x+(32)2 However the right hand side is not yet the same as the originalbracketed expression(x2 3x+12).To make it the same we subtract(32)2from both sides(x 32)2 (32)2=x2 3x+ (32)2 (32)2and then add12to both sides:(x 32)2 (32)2+12=x2 3x+12 Sox2 3x+12=(x 32)2 (32)2+12=(x 32)2 (32)2+12=(x 32)2 74So2x2 6x+ 1 = 2((x 32)2 74)We havecompleted the squarefor the quadratic function2x2 6x+ mathcentre 2009 The minimum value of the functionf(x)will be2 ( 74)= 72whenx= this Example we will consider a quadratic function for which the coefficient ofx2is the functionf(x) = 3 + 8x 2x2.

10 We operate in the same way as before, taking outthe factor multiplying + 8x 2x2= 2(x2 4x 32)We now deal with just the bracketed term as the two expressions:x2 2ax+a2andx2 4x 32 Clearly the coefficients ofx2in both expressions are the same - they match would like to match up the term 2axwith the term 4x. To do this note that2amust be4, so thata= 2. As before we note(x a)2=x2 2ax+a2 Witha= 2(x 2)2=x2 4x+ 4 This is not yet the same as the original expressionx2 4x 32. To make it the same we cansubtract 4 from each side:(x 2)2 4 =x2 4x+ 4 4and subtract32from each side:(x 2)2 4 32=x2 4x 32so thatx2 4x 32= (x 2)2 4 32= (x 2)2 112 Then multiplying both sides by 2in order to recover our original function3 + 8x 2x2= 2((x 2)2 112)We havecompleted the this case, whenx= 2the function will have its maximum value, and this will be mathcentre 2009 Exercises3.