3.3
Solving Quadratic Equations
by Graphing
GOAL
YOU WILL NEED
Use graphs to solve quadratic equations.
• graph paper
• graphing calculator
INVESTIGATE the Math
A model rocket is launched from the roof of a building.
The height, h(t), in metres, at any time, t, in seconds, is modelled by the
function h(t) 5 25t 2 1 15t 1 20.
?
When will the rocket hit the ground?
A.
Use a graphing calculator to graph the height function. Determine the
zeros, the axis of symmetry, and the vertex of this function.
B.
What is the rocket’s height when it hits the ground? Use this value to
write a quadratic equation you could solve to determine when the
rocket hits the ground.
C.
Substitute one of the zeros you determined in part A into the
quadratic equation you wrote in part B. Repeat this for the other zero.
What do you notice?
D.
What are the roots of the quadratic equation you wrote in part B?
E.
State the domain and range of the function in the context of
this problem.
F.
What is the starting height of the rocket? Where is it on the graph?
Where is it in the function?
G.
NEL
When will the rocket hit the ground?
quadratic equation
an equation that contains a
polynomial whose highest
degree is 2; for example,
x 2 1 7x 1 10 5 0
root of an equation
a number that when substituted
for the unknown, makes the
equation a true statement
for example, x 5 2
is a root of the equation
x 2 2 x 2 2 5 0 because
22 2 2 2 2 5 0
the root of an equation is also
known as a solution to that
equation
Working with Quadratic Functions: Standard and Factored Forms
143
Reflecting
H.
How are the zeros of the function h(t) related to the roots of the
quadratic equation 0 5 25t 2 1 15t 1 20?
I.
The quadratic equation 0 5 25t 2 1 15t 1 20 has two roots.
Explain why only one root is an acceptable solution in this situation.
How does your explanation relate to the domain of the function h(t)?
J.
How can the graph of the function f (x) 5 ax 2 1 bx 1 c help you
solve the quadratic equation ax 2 1 bx 1 c 5 0?
APPLY the Math
EXAMPLE
1
Connecting graphs to the solutions
of a quadratic equation
Determine the solutions of the quadratic equation x 2 2 8x 1 12 5 23 by
graphing. Check your solutions.
Nash’s Solution: Rearranging the Equation to Determine
the Zeros on the Corresponding Graph
x 2 2 8x 1 12 5 23
x 2 2 8x 1 15 5 0
g(x) 5 x 2 2 8x 1 15
I rearranged the equation so that it
was equal to zero. I used the
corresponding function,
g(x) 5 x 2 2 8x 1 15, to solve the
equation by finding its zeros.
I graphed g(x) 5 x 2 2 8x 1 15.
I used the zero operation on my
calculator to see where the graph
crosses the x-axis.
One solution is x 5 3.
There’s another point where the
graph crosses the x-axis, so I used
the zero operation again.
The other solution is x 5 5.
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Chapter 3
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3.3
Tech
Support
For help determining values
and zeros using a graphing
calculator, see Technical
Appendix, B-3 and B-8.
To check, I entered the left side of the
original equation (x 2 2 8x 1 12)
into Y1, and then used the value
operation to see if the y-value is
equal to 23.
Both solutions give a value of 23,
so I know they’re correct.
Rearranging a quadratic equation into the form ax 2 1 bx 1 c and then
graphing f (x) 5 ax 2 1 bx 1 c is one method you can use to solve a quadratic
equation. Another method is to treat each side of the equation as a function
and then graph both to determine the point(s) of intersection.
Lisa’s Solution: Determining the Points of Intersection
of Two Corresponding Functions
x 2 2 8x 1 12 5 23
f (x) 5 x 2 2 8x 1 12 and
g(x) 5 23
I separated the equation into two
functions, using each side of the
equation. I called the left side
f(x) and the right side g(x).
The solutions of the original
equation correspond to the
points that lie on both graphs at
the same time. I can determine
these solutions by graphing both
functions and locating the points
of intersection.
NEL
Tech
Support
For help determining points of
intersection using a graphing
calculator, see Technical
Appendix, B-11.
Working with Quadratic Functions: Standard and Factored Forms
145
I found the points of intersection by
using the intersect operation on the
calculator. The x-coordinates of
these points are the solutions of the
original equation.
The two solutions are x 5 3 and x 5 5.
Check:
L.S. 5 x 2 2 8x 1 12
R.S. 5 23
When x 5 3,
L.S. 5 (3) 2 2 8(3) 1 12
5 9 2 24 1 12
5 23
5 R.S.
When x 5 5,
I substituted each of my answers
into the original equation to see if I
had the correct solutions. They both
work, so I know my solutions are
correct.
L.S. 5 (5) 2 2 8(5) 1 12
5 25 2 40 1 12
5 23
5 R.S.
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Chapter 3
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3.3
EXAMPLE
2
Selecting strategies to make predictions from a quadratic equation
The population of an Ontario city is modelled by the function P(t) 5 0.5t 2 1 10t 1 300,
where P(t) is the population in thousands and t is the time in years. Note: t 5 0 corresponds to
the year 2000.
a) What was the population in 2000?
b) What will the population be in 2010?
c) When is the population expected to be 1 050 000?
Guillaume’s Solution
a) P(0) 5 0.5(0) 2 1 10(0) 1 300
The year 2000 corresponds to t 5 0, so I set t 5 0 in the
equation and evaluated. I remembered to multiply the
solution by 1000 because P is in thousands.
5 0 1 0 1 300
5 300
The population was 300 000 in the year 2000.
b) P(10) 5 0.5(10) 2 1 10(10) 1 300
The year 2010 corresponds to t 5 10, so I set t 5 10 in
the equation. Again, I had to remember to multiply the
solution by 1000 because P is in thousands.
5 0.5(100) 1 100 1 300
5 50 1 100 1 300
5 450
The population will be 450 000 in the year 2010.
To see when the population will be 1 050 000,
I let P be 1050 because P is in thousands, and
c) 1050 5 0.5t 2 1 10t 1 300
0 5 0.5t 2 1 10t 2 750
1 050 000
5 1050.
1000
I rearranged the equation to get zero on the left
side. I used my graphing calculator to graph
P(t) 5 0.5t 2 1 10t 2 750, and then I looked for the
zeros. I decided to use t $ 0 as the domain, so I’ll have
only one zero.
Using the zero operation on my graphing calculator, I got
30 for the zero.
The population should reach 1 050 000 in the
year 2030.
NEL
Since t 5 0 is used to represent the year 2000, I added 30
to get 2030.
Working with Quadratic Functions: Standard and Factored Forms
147
EXAMPLE
3
Solving problems involving a quadratic
equation
The function h(t) 5 2 1 50t 2 1.862t 2, where h(t) is the height in metres
and t is time in seconds, models the height of a golf ball above the planet
Mercury’s surface during its flight.
a) What is the maximum height reached by the ball?
b) How long will the ball be above the surface of Mercury?
c) When will it reach a height of 200 m on the way down?
Candace’s Solution
Tech
Support
a)
I graphed the equation
y 5 2 1 50t 2 1.862t 2 on my
graphing calculator. I used the
maximum operation on my
calculator to get the maximum
value.
For help determining
maximum and minimum values
using a graphing calculator,
see Technical Appendix, B-9.
The maximum height of the golf ball
is about 337.7 m.
When the ball reaches a height of
0 m, it’s no longer above the planet.
So I need to identify the zeros.
b)
The first zero is before the golf ball
has been hit. I want the second
zero.
Tech
Support
For help determining points
of intersection using a
graphing calculator, see
Technical Appendix, B-11.
The ball will be above the planet’s
surface for about 26.9 s.
c) 2 1 50t 2 1.862t 2 5 200
The ball will reach a height of 200 m
on its way down at about 22.0 s.
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Chapter 3
To find the time when the ball is at
a height of 200 m, I set h(t) equal
to 200. To solve this equation,
I graphed y 5 2 1 50t 2 1.862t 2
and y 5 200 on the same axes.
From the graph, I can see that
there are two points of
intersection. The second point is
the height on the way down, so I
found it using the intersection
operation on the calculator.
NEL
3.3
In Summary
Key Idea
• All quadratic equations can be expressed in the form ax 2 1 bx 1 c 5 0 using
algebraic techniques. They can be solved by graphing the corresponding
function f(x) 5 ax 2 1 bx 1 c. The zeros, or x-intercepts, of the function are
the solutions, or roots, of the equation.
Need to Know
• A quadratic equation is any equation that contains a polynomial whose
highest degree is 2. For example, x 2 1 8x 1 15 5 0.
• An alternative to solving ax 2 1 bx 1 c 5 d is to graph both y 5 ax 2 1 bx 1 c
and y 5 d. The solutions will be those points where the two functions intersect.
• You should substitute the solutions into the original equation to verify the result.
CHECK Your Understanding
1. Graph each function by hand. Then use the graph to solve the
corresponding quadratic equation.
a) g(x) 5 2x 2 1 5x 1 14 and 2x 2 1 5x 1 14 5 0
b) f (x) 5 x2 1 8x 1 15 and x 2 1 8x 1 15 5 0
2. Graph each function using a graphing calculator. Then use the graph
to solve the corresponding quadratic equation.
a) h(x) 5 x 2 2 x 2 20 and x 2 2 x 2 20 5 0
b) f (x) 5 x 2 2 5x 2 9 and x 2 2 5x 2 9 5 0
PRACTISING
3. For each function, write the corresponding quadratic equation whose
solutions are also the zeros of the function.
f(x)
a)
⫺8 ⫺4 0
⫺10
4
⫺20
⫺30
⫺40
y = x2 ⫺ 2x ⫺ 35
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x
8
f(x)
b)
8
4
⫺4 ⫺2 0
⫺4
x
2
4
⫺8
y = ⫺x2 + 3x + 4
Working with Quadratic Functions: Standard and Factored Forms
149
c)
f(x)
2
⫺6 ⫺4 ⫺2 0
⫺2
d)
x
2
⫺4
⫺6
y = x2 + 3x ⫺ 5
2
f(x)
⫺2 ⫺1 0
⫺2
x
1
2
⫺4
⫺6
⫺8
y = ⫺6x2 ⫺ x + 2
4. Graph the corresponding function to determine the roots of each
equation. Verify your solutions.
a) x 2 2 8x 5 216
b) 2x 2 1 3x 2 20 5 0
c) 25x 2 1 15x 5 10
d) x2 1 4x 5 8
e) x2 1 5 5 0
f ) 4x 2 2 64 5 0
5. Graph each function. Then use the graph to solve the
K
quadratic equation.
a) p(x) 5 3x 2 1 5x 2 2 and 3x 2 1 5x 2 2 5 0
b) f (x) 5 2x 2 2 11x 2 21 and 2x 2 2 11x 2 21 5 0
c) p(x) 5 8x 2 1 2x 2 3 and 8x 2 1 2x 2 3 5 0
d) f (x) 5 3x 2 1 x 1 1 and 3x 2 1 x 1 1 5 0
6. The population, P(t), of an Ontario city is modelled by the function
P(t) 5 14t 2 1 650t 1 32 000. Note: t 5 0 corresponds to the
year 2000.
a) What will the population be in 2035?
b) When will the population reach 50 000?
c) When was the population 25 000?
7. The function h(t) 5 2.3 1 50t 2 1.86t 2 models the height of an
arrow shot from a bow on Mars, where h(t) is the height in metres
and t is time in seconds. How long does the arrow stay in flight?
8. The height of an arrow shot by an archer is given by the function
h(t) 5 25t 2 1 18t 2 0.25, where h(t) is the height in metres and
t is time in seconds. The centre of the target is in the path of the arrow
and is 1 m above the ground. When will the arrow hit the centre of
the target?
9. The student council is selling cases of gift cards as a fundraiser.
The revenue, R(x), in dollars, can be modelled by the function
R(x) 5 225x 2 1 100x 1 1500, where x is the number of cases of
gift cards sold. How many cases must the students sell to maximize
their revenue?
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Chapter 3
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3.3
10. The Wheely Fast Co. makes custom skateboards for professional
A
riders. The company models its profit with the function
P(b) 5 22b2 1 14b 2 20, where b is the number of skateboards
produced, in thousands, and P(b) is the company’s profit, in
hundreds of thousands of dollars.
a) How many skateboards must be produced for the company to
break even?
b) How many skateboards does Wheely Fast Co. need to produce to
maximize profit?
Communication
Tip
A company’s break-even point is
the point at which the company
shows neither a profit nor a loss.
This occurs when the profit is
zero.
11. A ball is tossed upward from a cliff that is 40 m above water. The height
of the ball above the water is modelled by h(t) 5 25t 2 1 10t 1 40,
where h(t) is the height in metres and t is the time in seconds. Use a
graph to answer the following questions.
a) What is the maximum height reached by the ball?
b) When will the ball hit the water?
12. The cost, C(n) , in dollars, of operating a concrete-cutting machine is
modelled by C(n) 5 2.2n2 2 66n 1 655, where n is the number of
minutes the machine is in use.
a) How long must the machine be in use for the operating cost to be
a minimum?
b) What is the minimum cost?
13. a) For each condition, determine an equation in standard form of a
T
quadratic function that
i) has two zeros
ii) has one zero
iii) has no zeros
b) What is the maximum number of zeros that a quadratic function
can have? Explain.
14. a) What quadratic function could be used to determine the solution
C
NEL
of the quadratic equation 3x 2 2 2x 1 5 5 4?
b) Explain how you could use the function in part (a) to determine
the solutions of the equation.
Working with Quadratic Functions: Standard and Factored Forms
151
Extending
15. A parabolic arch is used to support a bridge. Vertical cables are every
2m
2 m along its length. The table of values shows the length of the cables
with respect to their placement relative to the centre of the arch.
Negative values are to the left of the centre point. Write an algebraic
model that relates the length of each cable to its horizontal placement.
Distance from Centre
of Arch (m)
Length of Cable
(m)
210
120.0
28
130.8
26
139.2
24
145.2
22
148.8
0
150.0
2
148.8
4
145.2
6
139.2
8
130.8
10
120.0
16. Determine the points of intersection of
a) y 5 4x 2 1 and y 5 2x 2 1 5x 2 7
b) y 5 x 2 2 2x 2 9 and y 5 2x 2 1 5x 1 6
17. Show that the function y 5 2x 2 2 3x 1 4 cannot have a y-value less
than 1.5.
152
Chapter 3
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