Rewriting a Quadratic Expression

Task

1. What is the minimum value taken by the expression $(x-4)^2 + 6$? How does the structure of the expression help to see why?
2. Rewrite the quadratic expression $x^2 - 6x - 3$ in the form $(x-\underline{\hspace{.5cm}})^2+\underline{\hspace{.5cm}}$ and find its minimum value.
3. Rewrite the quadratic expression $-2x^2 + 4x + 3$ in the form $\underline{\hspace{.5cm}}(x-\underline{\hspace{.5cm}})^2+\underline{\hspace{.5cm}}$. What is its maximum value? Explain how you know.

IM Commentary

The goal of this task is to complete the square in a quadratic expression in order to find its minimum or maximum value. When we have a quadratic function $f(x)$ this corresponds to finding the vertex of its graph. Completing the square in a quadratic expression helps to find when the expression is equal to 0. For example, working with the expression from part (b), we have $$x^2 - 6x - 3 = (x-3)^2 - 12.$$ We then set this equivalent expression equal to 0, $$(x-3)^2 - 12 = 0,$$ and solve by adding 12 to both sides of the equation and then extracting a square root. When applied to a general quadratic expression $ax^2 + bx + c$, completing the square leads to the quadratic formula.

Solution

1. The quantity $(x-4)^2$ is never negative since it is the square of a real number. We have $(x-4)^2 = 0$ only when $x - 4 = 0$. So the expression $(x-4)^2$ has a minimum value of 0 when $x = 4$. This means that the expression $(x-4)^2 + 6$ has a minimum value when $x = 4$ and that minimum value is 6.

   The structure of the expression $(x-4)^2 + 6$ was vital in determining its minimum value as it allows us to focus on the simpler expression $(x-4)^2$ and use the fact that the only real number whose square equals 0 is 0.

2. In order to write $x^2 - 6x - 3$ as a perfect square plus a number, we focus on the first two terms $x^2 - 6x$. To write $x^2 - 6x$ as a perfect square plus a number note that, for any number $a$, $$(x+a)^2 = x^2 + 2ax +a^2.$$ We have $-6x$ in the expression $x^2 - 6x$ so this means we want $2a = -6$ or $a = -3$. With this choice of $a$ we find $$x^2 - 6x - 3 = (x-3)^2 -12.$$

   As in part (a), the minimum value will occur when $(x-3)^2 = 0$ or $x = 3$. When $x = 3$ we see that the minimum value is -12.

3. For $-2x^2 + 4x + 3$ we can work as in parts (a) and (b) but it is convenient to first factor out the leading coefficient of -2: $$-2x^2 + 4x + 3 = -2\left(x^2 - 2x - \frac{3}{2}\right).$$ Alternatively, we could write $-2x^2 + 4x +3 = -2(x^2 -2x) + 3$. For the expression $x^2 -2x - \frac{3}{2}$ we focus on $x^2 - 2x$ and, working as in part (b), we will want to choose $a = -1$ giving $$x^2 - 2x - \frac{3}{2} = (x-1)^2 - \frac{5}{2}.$$ Substituting this into the above equality gives $$-2x^2 + 4x + 3 = -2\left((x-1)^2 - \frac{5}{2}\right).$$ With the alternate expression, we would find $-2x^2 +4x +3 = -2(x-1)^2+5$, the expanded form of the right hand side of this equation (and the form requested in the question). The expression $-2(x-1)^2+5$ takes a maximum value when $(x-1)^2 = 0$ or $x = 1$. When $x = 1$, this maximum value is $5$. We know that the value is a maximum because if $x \neq 1$ then $(x-1)^2 \gt 0$ and so $-2(x-1)^2 \lt 0$ and the value of the expression is less than $5$.