8.4 Applications of Quadratic Functions

SL 2.6

In this section you will learn

how quadratic functions model real situations, such as the path of a ball or the shape of an arch
how to answer “greatest height”, “when” and “does it hit / does it fit” questions with the vertex and with substitution
how to build a quadratic model from information about a shape

Quadratic models

Quadratic functions model many practical situations:

the path (trajectory) of a ball or a jet of water;
parabolic structures such as arches, bridges and satellite dishes;
areas, where one length grows while another shrinks;
profit or revenue, which rise and then fall.

The vertex gives the greatest or least value. The intercepts give the start and the end.

Working with a quadratic model

Read carefully what each variable means, and its units.
“Greatest” or “least” value: find the vertex (complete the square, or use $x = -\frac{b}{2a}$).
“At the start”, “when $x = \ldots$”: substitute into the function.
“When does it reach $\ldots$”: solve an equation.
Answer in words, in the context, with units. Check that the answer makes sense.

Example 1 A stone is thrown from the top of a small wall. Its height, $h$ m, above the ground is given by

\[ h = -0.5x^2 + 3x + 1.5, \]

where $x$ m is the horizontal distance from the wall.

(a) Find the height of the stone at the moment it is thrown.

(b) Find the greatest height of the stone.

(c) A target is $4$ m horizontally from the wall and $5$ m above the ground. Determine whether the stone hits the target.

Solution

(a) At the moment it is thrown, $x = 0$: $\;h = 1.5$ m.

(b) Complete the square:

\[ -0.5x^2 + 3x + 1.5 = -0.5(x^2 - 6x) + 1.5 = -0.5\big[(x - 3)^2 - 9\big] + 1.5 = -0.5(x - 3)^2 + 6 . \]

The greatest height is $6$ m, when the stone is $3$ m from the wall.

(c) At $x = 4$: $\;h = -0.5(16) + 12 + 1.5 = 5.5$ m.

The stone is $5.5$ m high at that point, not $5$ m. It passes $0.5$ m above the target, so it does not hit it.

Sometimes you must build the model yourself. Choose the axes so that the numbers are simple, then use the vertex form $y = a(x - h)^2 + k$.

Example 2 A tunnel entrance has the shape of a parabolic arch. The road under the arch is $8$ m wide, and the greatest height of the arch is $6$ m. Take the origin at one end of the arch, as in the diagram, with $x$ m the horizontal distance and $y$ m the height.

(a) Find a quadratic function in the form $y = a(x - h)^2 + k$ for the arch.

(b) Find the height of the arch at the points where its width is $4$ m.

(c) A lorry is $4$ m wide and $4.6$ m tall. It drives through the middle of the road. Decide whether it can pass under the arch.

Solution

(a) By symmetry the vertex is halfway across: $(4, 6)$. So $y = a(x - 4)^2 + 6$.

The arch passes through $(0, 0)$: $\;a(16) + 6 = 0$, so $a = -\dfrac{3}{8}$.

\[ y = -\frac{3}{8}(x - 4)^2 + 6 . \]

(b) A width of $4$ m, centred on $x = 4$, goes from $x = 2$ to $x = 6$.

\[ y = -\frac{3}{8}(2 - 4)^2 + 6 = -\frac{3}{8}(4) + 6 = 4.5 \text{ m}. \]

(c) The lorry, in the middle, fills $x = 2$ to $x = 6$. At its edges the arch is only $4.5$ m high, but the lorry is $4.6$ m tall. So the lorry cannot pass under the arch.

Paper 2 — models on the TI-Nspire CX II

Graphs page: enter the model as $f_1(x)$ and set a sensible window (only $x \geq 0$ and $y \geq 0$ usually make sense).

Greatest value: Menu → Analyze Graph → Maximum.
Height at a given $x$: Menu → Trace → Graph Trace, then type the $x$-value and press enter.
When the height is a given value: graph $f_2(x) = $ that value and use Menu → Analyze Graph → Intersection.

Common errors

✗ Wrong	✓ Right	Why
Greatest height at $x = 0$	At the vertex	$x = 0$ gives the starting height.
Answer “$(3, 6)$” for “greatest height”	“$6$ m”	Answer the question that was asked, with units.
Width $4$ m means $x = 4$	$x$ from $2$ to $6$	The width is measured across the axis of symmetry.
Using the model far outside the data	Comment on the limits	For example, a negative height is not possible.

Exercises

1 A ball is kicked from the ground. Its height after $t$ seconds is $h = 20t - 5t^2$ metres.

(a) Find the greatest height. $\qquad$ (b) Find when the ball lands.

Solution

(a) $h = -5(t^2 - 4t) = -5(t - 2)^2 + 20$. Greatest height $20$ m (at $t = 2$ s).

(b) $5t(4 - t) = 0$, so $t = 0$ or $t = 4$. It lands after $4$ s.

2 A farmer has $60$ m of fencing to make a rectangular pen. Let one side be $x$ m.

(a) Show that the area is $A = 30x - x^2$.

(b) Find the greatest possible area.

Solution

(a) The perimeter is $60$, so the other side is $30 - x$. $\;A = x(30 - x) = 30x - x^2$.

(b) $A = -(x - 15)^2 + 225$. The greatest area is $225\ \text{m}^2$ (a square of side $15$ m).

3 A rectangular garden is next to a wall. Fencing is used on the other three sides only, and there is $40$ m of fencing. Find the greatest possible area.

Solution

Let the two sides perpendicular to the wall be $x$ m. The side parallel to the wall is $40 - 2x$.

$A = x(40 - 2x) = -2x^2 + 40x = -2(x - 10)^2 + 200$. The greatest area is $200\ \text{m}^2$ (when $x = 10$).

4 A shop sells $200 - 4p$ caps per week when the price is $$p$. Find the price that gives the greatest weekly revenue, and that revenue.

Solution

$R = p(200 - 4p) = -4p^2 + 200p = -4(p - 25)^2 + 2500$.

Price $25 gives the greatest revenue, $2500.

5 The height of a model rocket is $h = -4.9t^2 + 29.4t$ metres, $t$ seconds after launch. Find its greatest height and the time when it reaches it.

Solution

$t = -\dfrac{29.4}{2(-4.9)} = 3$ s. $\;h = -4.9(9) + 29.4(3) = -44.1 + 88.2 = 44.1$ m.

6 The cost, $$C$, of making $n$ items per day is $C = 0.5n^2 - 40n + 1000$. Find the number of items that gives the least cost, and that cost.

Solution

$C = 0.5(n^2 - 80n) + 1000 = 0.5(n - 40)^2 - 800 + 1000 = 0.5(n - 40)^2 + 200$.

Least cost $200, when $40$ items are made.

7 The path of a jet of water from a fountain is $y = -0.8x^2 + 3.2x$, where $x$ m is the horizontal distance and $y$ m the height.

(a) How far from the fountain does the water land?

(b) What is the greatest height?

Solution

(a) $-0.8x(x - 4) = 0$: $x = 0$ or $x = 4$. It lands $4$ m away.

(b) At $x = 2$: $\;y = -3.2 + 6.4 = 3.2$ m.

8 A parabolic bridge support is $20$ m wide at road level and $5$ m high in the middle. Taking the origin at the top, write its equation in the form $y = ax^2$, and find the height of the support $4$ m from the centre.

Solution

The ends are at $(\pm 10, -5)$. $\;-5 = a(100)$, so $a = -\dfrac{1}{20}$ and $y = -\dfrac{1}{20}x^2$.

At $x = 4$: $y = -\dfrac{16}{20} = -0.8$, so the support is $0.8$ m below the top, that is, $5 - 0.8 = 4.2$ m above the road.

9 A ball’s height is $h = -t^2 + 6t + 1$ metres after $t$ seconds. For how long is the ball higher than $6$ m?

Solution

$-t^2 + 6t + 1 = 6 \Rightarrow t^2 - 6t + 5 = 0 \Rightarrow (t - 1)(t - 5) = 0$.

The ball is above $6$ m between $t = 1$ and $t = 5$, so for $4$ seconds.

10 The stretch, $s$ mm, of a spring with a load of $m$ g is modelled by $s = 0.001m^2 + 0.05m$, for $0 \leq m \leq 300$.

(a) Find the stretch for a load of $150$ g.

(b) Explain why the model should not be used to predict the stretch for a load of $5$ kg.

Solution

(a) $s = 0.001(22\,500) + 0.05(150) = 22.5 + 7.5 = 30$ mm.

(b) $5$ kg $= 5000$ g is far outside the range $0 \leq m \leq 300$ used to make the model. The model would give $s = 25\,250$ mm, about $25$ m, which is not realistic — the spring would break or behave differently.