Completing the Squares

Figuring out how to find the turning point

3-minute read
Table of Contents

A quadratic polynomial is a polynomial whose degree (the highest power of $x$) is $2$. The general form of a quadratic polynomial is:

$$ \begin{equation}\begin{aligned} y=ax^2+bx+c\\ \end{aligned}\end{equation} $$

The y-intercept

From this equation, we let $x=0$ to get the y-value of the y-intercept: $$ \begin{equation}\begin{aligned} y&=a(0)^2+b(0)+c\\ y&=c\\ \end{aligned}\end{equation} $$

Hence the y-intercept is the point $(0,c)$.

The x-intercepts a.k.a. the roots of the curve

We can also find the x-intercepts of the curve if we let $y=0$:

$$ \begin{equation}\begin{aligned} 0=ax^2+bx+c\\ \end{aligned}\end{equation} $$

This can be solved by factorization or the quadratic formula:

$$ \begin{equation}\begin{aligned} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \end{aligned}\end{equation} $$

Knowing these points intercepts are important in sketching the curve but we still have the issue of determining where the curve is supposed to turn around.

The turning point of the curve

If we can write the curve in the form:

$$ \begin{equation}\begin{aligned} a(x+h)^2+k\\ \end{aligned}\end{equation} $$

Then the coordinates of the turning point of the curve is simply $(-h,k)$.

The vertical line ($x=-h$) passing through the turning point is referred to as the axis of symmetry of the curve because the curve is symmetrical about this line.

What k means

The y coordinate of the turning point is known as the maximum/minimum value of the curve (based on the nature of the turning point).

For a maximum curve, the value $y=k$ is the maximum value. For a minimum curve, $y=k$ is the minimum value.

Steps in completing the square

Consider the quadratic expression: $$ \begin{equation}\begin{aligned} 2x^2+3x+5\\ \end{aligned}\end{equation} $$

Step 1 Pair the terms in $x$: $$ \begin{equation}\begin{aligned} (2x^2+3x)+5\\ \end{aligned}\end{equation} $$

Step 2 Factorize out the coefficient of $x^2$: $$ \begin{equation}\begin{aligned} 2(x^2+\frac{3}{2}x)+5\\ \end{aligned}\end{equation} $$

Step 3 Add the square of half of the coefficient of $x$ to the bracket, subtracting the same from the outside: $$ \begin{equation}\begin{aligned} 2(x^2+\frac{3}{2}x\color{red}+(\frac{3}{4})^2\color{normal})+5\color{royalblue}-2\color{normal}(\color{red}\frac{3}{4}\color{normal})^2\\ \end{aligned}\end{equation} $$

Note that the coefficient of $x^2$ ($2$ in this example) matters as adding any value to the bracket implies the value is being multiplied by that coefficient ($2$).

Step 4 By perfect square factorization, condense the contents of the bracket: $$ \begin{equation}\begin{aligned} 2(x+\frac{3}{4})^2+5-2(\frac{3}{4})^2\\ \end{aligned}\end{equation} $$

Step 5 Simplify the final terms: $$ \begin{equation}\begin{aligned} &2(x+\frac{3}{4})^2+5-2(\frac{9}{16})\\ &2(x+\frac{3}{4})^2+\frac{5}{1}-\frac{9}{8}\\ &2(x+\frac{3}{4})^2+\frac{40-9}{8}\\ &2(x+\frac{3}{4})^2+\frac{31}{8}\\ \end{aligned}\end{equation} $$

We thus have a turning point of $(-\frac34, \frac{31}{8})$. This means that the axis of symmetry is $x=-\frac{3}{4}$ and the minimum value of the curve will be $\frac{31}{8}$.

Test your knowledge

Complete the square for each the following:

  • $x^2+5x+6$
  • $3x^2-7x+1$
  • $-2x^2+9x-12$
  • $-x^2-5x$

Research Questions

Upon completion of this section you will be competent enough to answer the following:

  • What is a polynomial?
  • What is meant by the degree of a polynomial?
  • What is a quadratic polynomial?
  • What does $c$ represent in the equation $y=ax^2+bx+c$?
  • How do we find the roots (the x-intercepts) of a quadratic curve?
  • How do we complete the square to transform $ax^2+bx+c$ into $a(x+h^2)+k$?
  • What is the geometric significance of using the form $a(x+h)^2+k$ of a quadratic curve?
  • What are the coordinates of the turning point if given $a(x+h)^2+k$?
  • What is the axis of symmetry of a quadratic curve?

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