Year 10 Maths - 10 Parabolas and Other Graphs

Applications of Parabolas

20 practice questions 0 video lessons Theory + worked examples

Every question with a fully worked solution.

Theory

Many real situations follow a quadratic model: the height of a thrown ball, the area enclosed by a fixed length of fencing, the shape of an arch, or a business's profit. The turning point gives the greatest or least value, and the zeros show when the quantity is zero. This Year 10 maths page shows how to build, read and solve these models, and how to reject answers that do not fit the situation. Curriculum: VC2M10A15 / AC9M10A04.

A quadratic model is a rule such as $h = at^2 + bt + c$ that describes a real quantity. Its graph is a parabola, but usually only part of it makes sense: time, length and price cannot be negative.

The turning point gives the maximum (if $a \lt 0$) or minimum (if $a \gt 0$) value of the quantity, and the input that produces it.

The zeros (solutions of $\ldots = 0$) give when or where the quantity is zero: a ball landing, a profit of nothing (break-even), an arch meeting the ground.

Solutions that do not fit the situation, such as a negative time or a negative speed, are rejected, with a reason.

Often you build the model yourself first: name a variable, write the other quantities in terms of it, then multiply or add to get the quadratic.

For a thrown object: the turning point is the greatest height, the $h$-intercept is the starting height, and the positive zero is when it lands.

The maximum or minimum is on the axis of symmetry, halfway between the zeros.

The greatest or least value of $y = ax^2 + bx + c$ occurs on the axis of symmetry:

\[x = -\dfrac{b}{2a}\]

x = - \frac{b}{2 a}

If the model is in factored form $y = k(x - p)(x - q)$, the axis is halfway between the zeros:

\[x = \dfrac{p + q}{2}\]

x = \frac{p + q}{2}

When a quadratic does not factorise, use the quadratic formula:

\[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Turning-point form models. In \(h = -5(t - 1.5)^2 + 12\) the greatest height \(12\) occurs at \(t = 1.5\) straight away. Substitute \(t = 0\) for the starting value.

Between or outside? For a concave down model, the values above a level lie between the two solutions; for a concave up model, the values below a level lie between them.

How to solve a quadratic model problem

Build the model if it is not given: name the variable, write the other quantities in terms of it, then multiply or add.
Greatest or least value: find the axis $x = -\dfrac{b}{2a}$ (or the midpoint of the zeros), then substitute.
When does it equal a value? Set the model equal to the value, rearrange to $\ldots = 0$, and factorise or use the formula.
Interpret: reject impossible solutions, give units, and answer in the words of the question.

Problem types

Projectiles: $h = 0$ gives the landing time; the positive root only. A "time after the top" is landing time minus time of greatest height.
Area and number problems: two lengths with a fixed total, or two numbers with a fixed sum, give $x(L - x)$ type models whose maximum is halfway between the zeros.
Arches and bridges: place the origin at one foot (or the centre), use the turning-point form with the top as vertex, and substitute a foot to find $a$. For the width at a given height, set $y$ equal to that height and find the distance between the two solutions.
Vertical distance between two curves: subtract the lower expression from the upper one, simplify to a new quadratic, and find its maximum.
Finding coefficients: substitute given data points to get equations in $a$, $b$, $c$; $t = 0$ usually gives $c$ at once. A known turning point at $x = k$ gives $-\dfrac{b}{2a} = k$.
Price and quantity (revenue): if each $\$1$ rise in a $\$10$ price loses $2$ of $60$ sales, then after $n$ rises the revenue is $R = (10 + n)(60 - 2n)$, which expands to $R = 600 + 40n - 2n^2$, greatest at $n = 10$: a price of $\$20$ and $40$ sales.
Least distance: distance involves a square root, so minimise $D^2$ (a quadratic) instead, then take the square root.
Decimals in the model: multiply through by $10$ or $100$ before factorising, e.g. $0.01v^2 + 0.4v - 45 = 0$ becomes $v^2 + 40v - 4500 = 0$.

Example 1 — A ball thrown upwards

A ball is thrown upwards from a balcony. Its height above the ground after $t$ seconds is $h = 1.5 + 14t - 5t^2$ metres.
(i) Find its greatest height and when it occurs.
(ii) When does it hit the ground? Answer correct to $2$ decimal places.

Solution

(i) $a = -5 \lt 0$, so the turning point is a maximum. It lies on the axis of symmetry:

$t$	$=$	$-\dfrac{b}{2a} = -\dfrac{14}{2(-5)} = 1.4$
$h$	$=$	$1.5 + 14(1.4) - 5(1.4)^2$
	$=$	$1.5 + 19.6 - 9.8 = 11.3$

t = - \frac{b}{2 a} = 1.4

Greatest height $\mathbf{11.3}$ m, after $\mathbf{1.4}$ s.

(ii) Put $h = 0$: $5t^2 - 14t - 1.5 = 0$. It does not factorise, so use the quadratic formula with $a = 5$, $b = -14$, $c = -1.5$:

$t$	$=$	$\dfrac{14 \pm \sqrt{196 + 30}}{10}$
	$=$	$\dfrac{14 \pm \sqrt{226}}{10}$
$t$	$\approx$	$2.90\ \text{or}\ -0.10$

t = \frac{14 \pm \sqrt{226}}{10}

Time cannot be negative, so the ball hits the ground after $\mathbf{2.90}$ s.

Example 2 — Greatest area

A farmer has $120$ m of fencing to make a rectangular paddock split into two pens by a fence parallel to one side, as shown. Find the dimensions that give the greatest total area, and that area.

Solution

Let the three parallel fences each be $x$ m and the total length be $2y$ m. The fencing used is $3x + 2(2y) = 120$, so $2y = 60 - 1.5x$.

$A$

$=$

$x(60 - 1.5x)$

A = x (60 - 1.5 x)

The zeros are $x = 0$ and $x = 40$, and the parabola is concave down. The maximum is halfway between the zeros:

$x$	$=$	$20$
$\text{length}$	$=$	$60 - 1.5(20) = 30$
$A$	$=$	$20 \times 30 = \mathbf{600}\text{ m}^2$

The paddock is $\mathbf{20}$ m by $\mathbf{30}$ m. Check the fencing: $3(20) + 2(30) = 120$.

Example 3 — Will the truck fit?

A tunnel entrance is a parabolic arch $8$ m wide at road level and $6$ m high in the middle.
(i) Find its equation, taking the left foot as the origin.
(ii) Can a truck $3$ m wide and $4.8$ m tall pass through if it drives down the centre?

Solution

(i) The turning point is $(4,\ 6)$, so $y = a(x - 4)^2 + 6$. Substitute the foot $(0,\ 0)$:

$0$	$=$	$16a + 6$
$a$	$=$	$-\dfrac{3}{8}$
$y$	$=$	$-\dfrac{3}{8}(x - 4)^2 + 6$

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

(ii) The truck's top corners are $1.5$ m either side of the centre, at $x = 2.5$ and $x = 5.5$. By symmetry, check one:

$y$	$=$	$-\dfrac{3}{8}(2.5 - 4)^2 + 6$
	$=$	$-\dfrac{3}{8}(2.25) + 6$
	$\approx$	$5.16\text{ m}$

The arch is about $5.16$ m high above each corner, more than $4.8$ m, so yes, the truck fits, with about $0.36$ m to spare.

Example 4 — When is the profit high enough?

A market stall's daily profit from selling candles at $\$x$ each is $P = -3x^2 + 210x - 2700$ dollars.
(i) Find the price for the greatest profit, and that profit.
(ii) For which prices is the profit at least $\$675$?

Solution

(i) Axis of symmetry:

$x$	$=$	$-\dfrac{210}{2(-3)} = 35$
$P$	$=$	$-3(35)^2 + 210(35) - 2700$
	$=$	$-3675 + 7350 - 2700 = 975$

Greatest profit $\mathbf{\$975}$ at a price of $\mathbf{\$35}$.

(ii) Find where the profit equals $675$:

$-3x^2 + 210x - 2700$	$=$	$675$
$-3x^2 + 210x - 3375$	$=$	$0$
$x^2 - 70x + 1125$	$=$	$0$
$(x - 25)(x - 45)$	$=$	$0$

(x - 25) (x - 45) = 0

The graph is concave down, so it is above $675$ between these values: $\boldsymbol{25 \le x \le 45}$, that is, prices from $\$25$ to $\$45$.

Common pitfalls

Answer with the right quantity. The axis of symmetry gives the input (time, width, price). Substitute it to get the greatest or least value itself.

Reject with a reason. A negative time or length is not just dropped: say why, e.g. "$t \ge 0$, so reject $t = -0.10$".

Signs in $-\dfrac{b}{2a}$. With $a = -5$ and $b = 14$, $-\dfrac{14}{2(-5)} = 1.4$, a positive number. Keep the brackets.

Use the full model width. For an arch, the width at a height is the distance between the two $x$-values, $2\sqrt{\ldots}$, not just one of them.

Frequently asked questions

How do you find the maximum height of a projectile?

Find the time on the axis of symmetry with t equals negative b over 2a, then substitute that time into the height formula. The result is the greatest height.

How do you find when a ball hits the ground?

Set the height equal to zero and solve the quadratic, by factorising or with the quadratic formula. Keep the positive solution, because time after the throw cannot be negative.

How do you find the maximum area with a fixed perimeter?

Write the area as one variable times an expression for the other side, which gives a quadratic. Its maximum is on the axis of symmetry, halfway between its two zeros.

Why do we reject some answers in quadratic problems?

A quadratic equation can have two solutions, but the situation may only allow one, for example a positive time, length or speed. State which solution is rejected and why.

How do you find the equation of a parabolic arch?

Use the highest point as the vertex in y = a(x - h) squared + k, then substitute the point where the arch meets the ground to find a.

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\(t\)	\(=\)	\(-\dfrac{b}{2a} = -\dfrac{14}{2(-5)} = 1.4\)
\(h\)	\(=\)	\(1.5 + 14(1.4) - 5(1.4)^2\)
\(\)	\(=\)	\(1.5 + 19.6 - 9.8 = 11.3\)

\(t\)	\(=\)	\(\dfrac{14 \pm \sqrt{196 + 30}}{10}\)
\(\)	\(=\)	\(\dfrac{14 \pm \sqrt{226}}{10}\)
\(t\)	\(\approx\)	\(2.90\ \text{or}\ -0.10\)

\(x\)	\(=\)	\(20\)
\(\text{length}\)	\(=\)	\(60 - 1.5(20) = 30\)
\(A\)	\(=\)	\(20 \times 30 = \mathbf{600}\text{ m}^2\)

\(0\)	\(=\)	\(16a + 6\)
\(a\)	\(=\)	\(-\dfrac{3}{8}\)
\(y\)	\(=\)	\(-\dfrac{3}{8}(x - 4)^2 + 6\)

\(y\)	\(=\)	\(-\dfrac{3}{8}(2.5 - 4)^2 + 6\)
\(\)	\(=\)	\(-\dfrac{3}{8}(2.25) + 6\)
\(\)	\(\approx\)	\(5.16\text{ m}\)

\(x\)	\(=\)	\(-\dfrac{210}{2(-3)} = 35\)
\(P\)	\(=\)	\(-3(35)^2 + 210(35) - 2700\)
\(\)	\(=\)	\(-3675 + 7350 - 2700 = 975\)

\(-3x^2 + 210x - 2700\)	\(=\)	\(675\)
\(-3x^2 + 210x - 3375\)	\(=\)	\(0\)
\(x^2 - 70x + 1125\)	\(=\)	\(0\)
\((x - 25)(x - 45)\)	\(=\)	\(0\)

Applications of Parabolas

📖 Theory

How to solve a quadratic model problem

Problem types

Common pitfalls

Frequently asked questions

Theory