Minus One Two Three
Substitute -1, -2 or -3, into an algebraic expression and you'll
get three results. Is it possible to tell in advance which of those
three will be the largest ?
Age
14 to 16
Challenge level
If you substitute $-1$, $-2$ or
$-3$, into an algebraic expression you'll get three results. Is it
possible to tell in advance which of those three substitutions will
produce the greatest result and which will produce the least
?
Substitute $-1$, $-2$ and $-3$ into $(x + 2)(x + 6)$ here, and see what you get.
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Now can you predict for these ?
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Look at the two brackets for the factors that make the product. You are evaluating it for $x$ values of -1, -2, and -3 . Think what you could change systematically about those factors and see how it affects the size of the product for each of the three $x$ values.
Some good thinking from Berny at Gordonstoun School, and others, connecting the algebra to the shape and position of the quadratic graph (parabola) :
When we set a problem on the NRICH site there's often an insight we hope you'll discover as you work your way around the problem.
In Minus One Two Three the insight was to see that expressions a. b. and c. below are just three different ways to represent the same quadratic form.
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b. is made from a., multiplying to 'remove the brackets', and
c. is decided by looking at b.
Deciding the c. form is sometimes called 'completing the
square', if you are not yet familiar with that, square $( x + 4 )$
to check that the three representations really do match.
It's, b, the second of those three forms, which helps most here because it reveals the axis of symmetry for the red graph below ( the blue graph is a simple x-squared parabola passing through the origin )
Looking at c. we can see that $x = -4$ will produce the lowest possible value for this expression ( $-4$ ).
The '$+ 4$' in the expression works like a boost for the $x$ value and lets everything on the $x$-squared curve happen $4$ 'earlier' - that is to the left of the origin.
Subtracting $4$, after the 'squaring', has the effect of lowering all values, or points on the graph, by $4$.
So c. is like the $x$-squared parabola but shifted four left and four down.
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For x smaller than $-4$ the graph will be 'falling' towards this minimum value and to the right (greater than $-4$) the graph will 'climb' again.
So as x values move from $-3$ , to $-2$ , to $-1$ , the value of the expression will be increasing, away to the right of that minimum point of $x = -4$.
So $-3$ will produce the lowest of these three values, and $-1$ will produce the highest.
In a similar way with the other expressions :
$(2x + 7)(x + 1)$ produces middle, lowest, and highest of the three values when $x = -3$, $-2$, and $-1$
$(2x - 3)(x + 5)$ produces highest, lowest, and middle of the three values when $x = -3$, $-2$, and $-1$
$(x - 3)(x - 1)$ produces highest, middle, and lowest of the three values when $x = -3$, $-2$, and $-1$
More generally, you might like to think about whether every pair of linear brackets, like a. would produce a parabola-type ($x$-squared) curve, and similarly whether every expression like b. would have to be symmetric about some $x$ value.
It would be good if students knew that:
- the product of a pair of linear brackets represents a quadratic form,
- quadratic forms are always symmetric and
- the axis of symmetry of a quadratic form is often worth locating.
As with many Stage 4 algebra-based problems, there is much to gain from students making up similar problems for each other to solve, and especially if group discussion allows appreciation of the different 'ways of seeing' to be shared.
It is always good to distinguish for students identities that are 'equal' because the algebra is equivalent, from situations where the equals sign invites answers to the question 'for what values of the variable will these two expressions be equal in value?' - solving the equation?