Lesson 4 of 4Quadratic Equations

Unit 5 · Lesson 4 of 4

Solving Problems with Quadratics

Answer real questions with quadratic relations: the maximum height of a thrown object, when it lands, how long it stays above a given height, and the dimensions or price that give the greatest area or revenue.

7 min read12 practice questionsAsk AI about this →

Key things to remember

1Vertical motion: $h = - 4.9 t^{2} + v_{0} t + h_{0}$ ( $h$ in metres, $t$ in seconds, air resistance ignored).
2The maximum or minimum value is the second coordinate of the vertex; it occurs at $x = - \frac{b}{2 a}$ , which is halfway between the zeros when the parabola has two.
3"When does it reach height $H$ ?" means solve $h = H$ . Reject roots outside the domain, such as negative times.
4For a parabola that opens down, the height is above $H$ between the two roots of $h = H$ .
5For maximum area or revenue, write the quantity as a product of two factors, find the zeros, and go halfway between them.

You toss a ball straight up from a height of 1.5 m at 14 m/s. Three natural questions follow. How high does it go? When does it hit the ground? And if a friend on a balcony 8 m up wants to catch it, for how long is the ball above 8 m? The height is a quadratic relation of time, $h = - 4.9 t^{2} + 14 t + 1.5$ , and each question asks about a different feature of its parabola. The skill in this lesson is translating each question into the right feature, then using the tools from this unit and the last one to find it.

Question	Feature of the parabola	How to find it
How high does it go?	The h-coordinate of the vertex (the maximum value)	Vertex form, or substitute $t = - \frac{b}{2 a}$
When is it highest?	The t-coordinate of the vertex	$t = - \frac{b}{2 a}$ , or halfway between the zeros
When does it land?	The positive zero (t-intercept)	Solve $h = 0$
When is it at height H?	Where the parabola meets the horizontal line h = H	Solve $h = H$
How high was it at the start?	The h-intercept	Substitute $t = 0$

Translating questions about a thrown object into features of its graph

Modelling a vertical throw

Near Earth's surface, the height of an object thrown straight up (ignoring air resistance) is modelled by $h = - 4.9 t^{2} + v_{0} t + h_{0}$ , where $h$ is in metres and $t$ in seconds. The starting height is $h_{0}$ (read "h nought"), the upward starting speed is $v_{0}$ in metres per second, and the $- 4.9$ comes from gravity: it is half of $9.8$ m/s², the rate at which gravity changes the object's speed. Because $a = - 4.9$ is negative, the parabola opens down, so it has a maximum. The model only describes the object while it is in the air, so the sensible domain runs from $t = 0$ to the moment it lands.

Fig 1Graph of h = −4.9t² + 14t + 1.5 from the throw to the landing. It starts at 1.5 m, peaks at the vertex (1.43, 11.5) and lands at about 2.96 s. The horizontal line h = 8 meets the curve at about 0.58 s and 2.27 s, and the time interval between them is shaded.

Here is the ball, question by question, with times rounded to the nearest hundredth of a second. Highest point. The axis of symmetry is $t = - \frac{b}{2 a} = - \frac{14}{2 ( - 4.9 )} = \frac{14}{9.8} \approx 1.43$ s. Substituting the unrounded value gives $h = 11.5$ m, the maximum height. Landing. Solve $- 4.9 t^{2} + 14 t + 1.5 = 0$ : $b^{2} - 4 a c = 196 + 29.4 = 225.4$ , and $t = \frac{- 14 \pm 225.4}{- 9.8}$ , which is about $- 0.10$ s or $2.96$ s. The negative root is a time before the throw, so reject it: the ball lands after about 2.96 s. Above 8 m. Solve $- 4.9 t^{2} + 14 t + 1.5 = 8$ , which is $- 4.9 t^{2} + 14 t - 6.5 = 0$ : $b^{2} - 4 a c = 196 - 127.4 = 68.6$ , and $t = \frac{- 14 \pm 68.6}{- 9.8}$ , about $0.58$ s or $2.27$ s. The parabola opens down, so the ball is above 8 m between those times: for $0.58 < t < 2.27$ , which is about $2.27 - 0.58 = 1.69$ s.

Four ways to find a maximum or minimum

The maximum or minimum value of a quadratic relation is the second coordinate of its vertex. To find the vertex of

y = a x^{2} + b x + c

: complete the square to get vertex form

y = a (x - h)^{2} + k

; or find the zeros and take the value halfway between them; or use

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

, the centre of the quadratic formula; or graph it with technology and read the vertex. Then substitute that

x

into the relation to get the maximum (if

a < 0

) or minimum (if

a > 0

The tall Jet d'Eau fountain shooting water high above Lake Geneva with a rainbow in its spray — PhotoGeneva's Jet d'Eau shoots water about 140 m into the air; ignoring air resistance, the water's height over time follows the same kind of quadratic model as a ball thrown straight up.Asmus Koefoed, CC BY-SA 3.0, via Wikimedia Commons

VideoUse a height model to find the maximum height, the time it is reached, and when the object hits the ground.Watch on YouTube

Maximum area

A farmer has 40 m of fencing to build a rectangular pen against the long wall of a barn, so the wall forms one side and the fence forms the other three. Let $x$ be the width of the two sides that meet the wall. The fence uses $2 x$ for those, leaving $40 - 2 x$ for the side parallel to the wall. The area is $A = x (40 - 2 x)$ . This is already in factored form, so its zeros are easy: $A = 0$ when $x = 0$ or $40 - 2 x = 0$ , that is, $x = 20$ . Both zeros describe a pen with no area. The maximum is on the axis of symmetry halfway between them, at $x = \frac{0 + 20}{2} = 10$ . Then the long side is $40 - 20 = 20$ m and the area is $10 \times 20 = 200$ m². The sensible domain is $0 < x < 20$ .

VideoWrite the area of a fenced rectangle as a quadratic and use its vertex to find the dimensions that give the largest area.Watch on YouTube

Maximum revenue

A school club sells 300 concert tickets at $20 each, for revenue of $6000. A survey suggests that each $1 increase in price would mean 10 fewer tickets sold. Let $x$ be the number of $1 increases. Then the price is $20 + x$ dollars, the number sold is $300 - 10 x$ , and revenue = price × number sold, so $R = (20 + x) (300 - 10 x)$ . The zeros are $x = - 20$ (the price would be $0) and $x = 30$ (at $50 nobody buys). Halfway between them is $x = \frac{- 20 + 30}{2} = 5$ . So the best price is $20 + 5 = 25$ dollars, the club sells $300 - 50 = 250$ tickets, and the maximum revenue is $25 \times 250 = 6250$ dollars, $250 more than now. Checking neighbours confirms it: 4 increases give $24 \times 260 = 6240$ dollars and 6 give $26 \times 240 = 6240$ dollars.

Solving for a given value

Some problems give the output and ask for the input. A rectangular lawn 20 m by 12 m is surrounded by a path of uniform width $x$ metres, and the lawn and path together cover 384 m². The outer rectangle is $(20 + 2 x)$ by $(12 + 2 x)$ , because the path adds $x$ on both ends of each side. So $(20 + 2 x) (12 + 2 x) = 384$ . Expand: $240 + 40 x + 24 x + 4 x^{2} = 384$ , so $4 x^{2} + 64 x - 144 = 0$ . Divide every term by 4: $x^{2} + 16 x - 36 = 0$ , which factors as $(x + 18) (x - 2) = 0$ . A width cannot be negative, so $x = 2$ m. Check: $24 \times 16 = 384$ .

Read the problem and define the variable, with units.
Write the relation (or use the one given), and the domain that makes sense.
Decide what the question asks for: a maximum or minimum (find the vertex), or the input that gives a certain output (solve an equation).
Solve with the most efficient method: factoring, the quadratic formula, completing the square, or graphing technology.
Check each root against the domain and reject any that make no sense.
Answer the question in a sentence, with units and the rounding asked for.

Answer the question that was asked

"How high?" wants the height (11.5 m), not the time it happens (1.43 s). "For how long?" wants a length of time (1.69 s), while "over what time interval?" wants the start and end times (from 0.58 s to 2.27 s). "What price?" wants the price ($25), not the number of increases (

x = 5

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

A projectile is fired straight up from a 1.5 m platform. Its height in metres after

t

seconds is

h = - 4.9 t^{2} + 50 t + 1.5

. Find its maximum height, when it hits the ground, and the time interval during which it is higher than 3 m. Round times to the nearest hundredth of a second.

Example 2

A farmer has 40 m of fencing for a rectangular pen against a barn wall (the wall is one side). Find the dimensions that give the largest area, and that area.

Example 3

300 tickets sell at $20 each. Each $1 price increase means 10 fewer tickets sold. What price gives the greatest revenue, and what is that revenue?

Example 4

A ball's height is

h = - 4.9 t^{2} + 14 t + 1.5

metres after

t

seconds. For how long is the ball higher than 8 m? Round to the nearest hundredth of a second.

Where marks get lost

Giving the time of the maximum as the maximum height. For $h = - 4.9 t^{2} + 14 t + 1.5$ , the vertex is about $(1.43, 11.5)$ : the ball is highest at 1.43 s, and the maximum height is 11.5 m.
Keeping a negative time. When $h = 0$ gives $t \approx - 0.10$ or $t \approx 2.96$ , the negative root is before the throw and must be rejected.
Giving the wrong part of the time line. For a parabola that opens down, "above 8 m" is the interval between the two roots, not before the first or after the second.
Sign slips in $t = - \frac{b}{2 a}$ : with $b = 50$ and $a = - 4.9$ , $- \frac{50}{2 ( - 4.9 )} = + 5.10$ , not $- 5.10$ .
Ignoring the wall in a fencing problem: with a wall on one side, the fence covers $2 x + length$ , not $2 x + 2 \times length$ .

Flashcards 8

Vertical motion model

h = - 4.9 t^{2} + v_{0} t + h_{0}

, what is

h_{0}

Maximum height

When does the object land?

+ 4 more

Practice

A ball's height is

h = - 4.9 t^{2} + 14 t + 1.5

metres after

t

seconds. What is its maximum height?

For the same ball,

h = - 4.9 t^{2} + 14 t + 1.5

, when does it hit the ground? Round to the nearest hundredth of a second.

A projectile's height is

h = - 4.9 t^{2} + 50 t + 1.5

metres after

t

seconds. Find its maximum height, rounded to one decimal place.

For the ball

h = - 4.9 t^{2} + 14 t + 1.5

, over what time interval is it higher than 8 m? (Times in seconds, rounded to hundredths.)

For that same ball, for how many seconds is it higher than 8 m? Round to the nearest hundredth.

You have 60 m of fencing for a rectangular garden against a wall, so only three sides need fence. What is the largest possible area?

m²

A store sells 500 water bottles a month at $10 each. Each $1 increase in price means 20 fewer sales, and the same rate applies to part-dollar increases (a $0.50 increase means 10 fewer sales). What price, in dollars, gives the greatest revenue?

dollars

What is the minimum value of

y = 2 x^{2} - 12 x + 23

A 10 cm by 15 cm photo has a frame of uniform width

x

cm around it. The photo and frame together cover 266 cm². Find

x

For a thrown ball modelled by

h = - 4.9 t^{2} + v_{0} t + h_{0}

, match each question with the feature of the graph that answers it.

What is the maximum height?

When is the ball highest?

When does the ball land?

How high was the ball when it was thrown?

Put the steps for finding when the ball

h = - 4.9 t^{2} + 14 t + 1.5

lands in order.

1State the answer: the ball lands after about 2.96 s
2Reject $t \approx - 0.10$ , because it is before the throw
3Use the formula: $t = \frac{- 14 \pm 225.4}{- 9.8}$ , so $t \approx - 0.10$ or $t \approx 2.96$
4Set $h = 0$ : $- 4.9 t^{2} + 14 t + 1.5 = 0$
5Calculate the discriminant: $196 + 29.4 = 225.4$

In the model

h = - 4.9 t^{2} + v_{0} t + h_{0}

, what does

h_{0}

represent? Answer in a few words.

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