Today's lesson
Parabolas model many real situations: the flight of a ball, the cable of a bridge, the area of a rectangle with a fixed perimeter. The key features we have been finding now mean something: the vertex is the maximum (or minimum), and the $x$-intercepts mark the start and finish.
Learning intentions
- Set up a quadratic model from a worded problem (define variables, form an equation)
- Read the meaning of key features in context: the vertex is the max/min, the intercepts are the start/landing
- Decide on a suitable range of values for the variable
- Sketch the model, labelling intercepts and the turning point
Part 1 — Reading a model (~8 min)
When applying quadratics we usually: define variables → form an equation → solve / find key features → decide a sensible range → sketch.
Key ideas
- The turning point gives the maximum or minimum value (e.g. greatest height, largest area).
- The $x$-intercepts are where the quantity is zero (e.g. when the projectile is at ground level).
- The $y$-intercept is the starting value (e.g. height when $d=0$).
- Choose a range that makes sense: distances, times and lengths cannot be negative.
Part 2 — A given model (Example 1, ~12 min)
Turning point (from turning point form): $(10,9)$ — a maximum.
$h$-intercept ($d=0$): $h=-\tfrac{1}{16}(-10)^2+9=2.75$, so it leaves the hand at $2.75$ m.
$d$-intercepts ($h=0$): $0=-\tfrac{1}{16}(d-10)^2+9 \Rightarrow (d-10)^2=144 \Rightarrow d-10=\pm12$, so $d=22$ (taking the positive value).
b Maximum height $=9$ m. c It travels $22$ m before landing.
Now you try: A ball is thrown upward with $h=20t-5t^2$ ($h$ in m, $t$ in s). Find the maximum height and how long until it lands. Answers: turning point $(2,20)$, so maximum height $20$ m; it lands when $h=0$, i.e. at $t=4$ seconds.
Part 3 — Forming a model (Example 2, ~14 min)
a $2\times\text{length}+2x=100$, so length $=50-x$.
b $A=\text{length}\times\text{width}=(50-x)x=50x-x^2$.
c Both sides must be positive: $0 e–f The $x$-intercepts of $A=x(50-x)$ are $x=0$ and $x=50$, so the turning point is halfway, at $x=25$; then $A=25\times25=625$. Maximum area $=625$ cm², with dimensions $25\text{ cm}\times25\text{ cm}$ (a square). Now you try: An 80 cm wire is bent into a rectangle of length $x$ cm. Then width $=40-x$, $A=x(40-x)$, $0
📺 Walkthrough: reading the javelin model $h=-\tfrac{1}{16}(d-10)^2+9$ — the vertex is the maximum height and the intercepts are where it leaves the hand and lands.
Practice 3.1 — read the model.
- A stone's height is $h=-(t-3)^2+9$ ($h$ in m, $t$ in s). When is it highest, and how high?
- For the same stone, when does it hit the ground ($h=0$, take the positive time)?
- A paddock has area $A=x(40-x)$. What value of $x$ gives the largest area, and what is it?
Part 4 — Quick quiz (5 min)
Pick the correct answer for each, then click Mark.
Q1. For a projectile, the maximum height is found at the parabola's:
Q2. For $h=20t-5t^2$, the ball returns to the ground ($h=0$) at:
Q3. A wire of 100 cm makes a rectangle of width $x$. Its area is:
Q4. The maximum area of $A=x(50-x)$ is:
Q5. For the javelin $h=-\tfrac{1}{16}(d-10)^2+9$, the height when it leaves the hand ($d=0$) is:
Working program — Cambridge Ex 7F (p628)
After the quiz, open Cambridge Chapter 7 (page 628) and complete the following:
| Set work | Extension |
|---|---|
| Questions 1–4, 6, 7, 9–11 | Questions 12–14 |
Define your variables clearly and state a sensible range for each worded problem.
Exit ticket — write in your book
- In a height-versus-time model, what does the turning point tell you?
- What do the $x$-intercepts mean for a thrown object?
- Why must the width of a rectangle satisfy $0