MCR3U1
U3L10
LINEAR–QUADRATIC SYSTEMS
PART A ~ INTRODUCTION
A linear function and a quadratic function can intersect at a maximum of two points.
y
y
y

x

x
x

TWO POINTS
ONE POINT
NO POINTS
The point(s) of intersection of a line and a parabola can be found:
 graphically
 algebraically
PART B ~ SOLVING LINEAR–QUADRATIC SYSTEMS
Example 
Solve the given system graphically:
y
𝑓 (𝑥) = −(𝑥 − 4)2 + 8
𝑔 (𝑥 ) = 𝑥 + 2
x
To determine the point(s) of intersection algebraically:
1. Use substitution to replace 𝑓(𝑥) in the quadratic function with
𝑔(𝑥) from the linear function (ie) equate the functions.
2. Solve for the variable (the x–coordinate of the POI).
3. Evaluate the linear function at this value (the y–coordinate of the POI).
MCR3U1
U3L10
Example 
Solve the given systems algebraically:
a)
𝑓 (𝑥) = 𝑥 2 + 5𝑥 − 7
𝑔(𝑥) = 6𝑥 + 5
b)
𝑓 (𝑥) = −𝑥 2 + 𝑥 + 3
𝑔(𝑥) = 7𝑥 + 12
Example 
Determine the number of point(s) of intersection of the given
quadratic and linear functions using the discriminant.
𝑓 (𝑥) = 3𝑥 2 − 2𝑥 − 1
𝑔(𝑥) = −𝑥 − 6
MCR3U1
U3L10
Example 
A skydiver jumped from an airplane and fell freely for some time
before releasing his parachute. His height, ℎ(𝑡), in metres, above
the ground at any time, t, in seconds is given by:
ℎ1 (𝑡 ) = −4.9𝑡 2 + 5000
ℎ2 (𝑡 ) = −4𝑡 + 4000
(before releasing parachute)
(after releasing parachute)
a)
Determine how long after jumping the parachute was released.
b)
Determine how high above the ground the skydiver was at that time.
HOMEWORK: p.198–199 #1a, 2a, 3, 4ac, 6, 8, 10, 11 (Think!), 13