Today's lesson
Completing the square turns any quadratic $y=x^2+bx+c$ into turning point form $y=(x-h)^2+k$. Once it's in that form you can read the turning point straight off, decide whether it's a maximum or a minimum, find the intercepts and sketch the curve.
Learning intentions
- Know that completing the square rewrites any quadratic in turning point form $y=a(x-h)^2+k$
- Read the turning point $(h,k)$ and axis of symmetry directly from that form
- Find the $y$-intercept (put $x=0$) and the $x$-intercepts (put $y=0$, then solve)
- Sketch the parabola with every key feature labelled
Part 1 — Completing the square (~8 min)
To make $x^2+bx$ a perfect square, add and subtract $\left(\tfrac{b}{2}\right)^2$. For example, to make $x^2+6x$ a perfect square we add $9$ (from $\left(\tfrac{6}{2}\right)^2$), since $x^2+6x+9=(x+3)^2$. So:
The turning point form $(x+3)^2-7$ tells us the vertex is at $(-3,-7)$ — a minimum.
Key ideas — turning point form $y=a(x-h)^2+k$
- The turning point is $(h,k)$. When $a>0$ it is a minimum; when $a<0$ it is a maximum.
- The axis of symmetry is $x=h$.
- Find the $y$-intercept by substituting $x=0$.
- Find the $x$-intercepts (if any) by substituting $y=0$ and solving — take the square root of both sides.
- To solve a perfect square such as $(x-1)^2=16$: $x-1=\pm4$, so $x=1\pm4$, giving $x=5$ or $x=-3$.
Part 2 — Reading key features (Example 1, ~10 min)
- i Turning point is a maximum at $(1,16)$ (because $a=-4<0$).
- ii $y$-intercept ($x=0$): $y=-4(0-1)^2+16=-4+16=12$, so $(0,12)$.
- iii $x$-intercepts ($y=0$): $0=-4(x-1)^2+16 \Rightarrow (x-1)^2=4 \Rightarrow x-1=\pm2$, so $x=3$ or $x=-1$. Intercepts $(-1,0)$ and $(3,0)$.
Now you try: For $y=-2(x+1)^2+18$ find the turning point, $y$-intercept and $x$-intercepts. Answers: maximum at $(-1,18)$; $y$-intercept $(0,16)$; $x$-intercepts $(-4,0)$ and $(2,0)$.
Part 3 — Sketching by completing the square (Example 2, ~14 min)
The plan: complete the square to find the turning point, find the $y$-intercept, find the $x$-intercepts (if any), then draw a smooth curve through them.
a) $y=x^2+6x+15 = (x+3)^2+6$. Turning point is a minimum at $(-3,6)$; $y$-intercept $(0,15)$. For $x$-intercepts: $0=(x+3)^2+6$ has no solution, so there are no $x$-intercepts (the whole curve sits above the $x$-axis).
b) $y=x^2-4x+2 = (x-2)^2-2$. Minimum at $(2,-2)$; $y$-intercept $(0,2)$. For $x$-intercepts: $0=(x-2)^2-2 \Rightarrow (x-2)^2=2 \Rightarrow x=2\pm\sqrt2$, i.e. $x\approx0.59$ and $x\approx3.41$.
Now you try: Sketch $y=x^2-2x+2$ (answer: minimum $(1,1)$, $y$-intercept $(0,2)$, no $x$-intercepts) and $y=x^2-3x-1$ (answer: minimum $\left(\tfrac32,-\tfrac{13}{4}\right)$, $y$-intercept $(0,-1)$, $x$-intercepts $x=\tfrac{3\pm\sqrt{13}}{2}$).
📺 Walkthrough: completing the square on $y=x^2-4x+2$, reading the turning point, then finding the intercepts and sketching.
Practice 3.1 — complete the square to write each in turning point form, then state the turning point.
- $y=x^2+2x-5$
- $y=x^2-6x+10$
- $y=x^2+4x+1$
- $y=x^2-5x+1$
b) $(x-3)^2+1$, TP $(3,1)$ (no $x$-intercepts)
c) $(x+2)^2-3$, TP $(-2,-3)$
d) $\left(x-\tfrac52\right)^2-\tfrac{21}{4}$, TP $\left(\tfrac52,-\tfrac{21}{4}\right)$
Part 4 — Quick quiz (5 min)
Pick the correct answer for each, then click Mark.
Q1. The turning point of $y=(x-4)^2+3$ is:
Q2. Completing the square, $x^2+8x$ becomes a perfect square by adding:
Q3. $y=(x+3)^2+6$ has how many $x$-intercepts?
Q4. $y=-2(x-1)^2+8$ has a turning point that is a:
Q5. Solving $(x-2)^2=2$ gives:
Working program — Cambridge Ex 7D (p615)
After the quiz, open Cambridge Chapter 7 (page 615) and complete the following:
| Set work | Extension |
|---|---|
| Questions 1–8 (½), 10 (½), 11 | Questions 9, 12 (⅓), 14 |
"(½)" means do every second part; "(⅓)" every third. Show the completing-the-square working for any sketch.
Exit ticket — write in your book
- What number do you add (and subtract) to complete the square on $x^2+10x$?
- From $y=(x-h)^2+k$, where is the turning point and how do you know if it is a max or a min?
- How can you tell from the turning point form that a parabola has no $x$-intercepts?