MCF3M Functions & Applications · Unit 3 — Quadratic Functions: Models and Forms
Lesson 3.3
Solving Quadratic Equations by Graphing
Read solutions from a graph, compare zeros with intersections, check by substitution, and choose meaningful answers in context.
Estimated time: 45 minutes
Learning goals
- explain roots, zeros and x-intercepts
- solve quadratic equations by graphing
- rearrange with zero on one side
- find intersections of two graphs
- check solutions by substitution
- interpret roots in context
- distinguish exact and approximate answers
Step 1 of 4
Warm Up
Success criteria
Step 2 of 4
Learn
See it — where does the curve meet zero?
Graphing method — zeros
x² − 5x + 6 = 0
Select all roots (x-values), or select none for no real roots.
A root is an x-value that makes an equation true. When a quadratic equals zero, roots are the x-values where its graph meets the x-axis.
Point (2, 0) → zero 2 → solution x = 2. Point (3, 0) → solution x = 3.
See it — solving by zeros
Graphing method — zeros
x² − 4x − 5 = 0
Select all roots (x-values), or select none for no real roots.
Worked example
Solve x² − 4x − 5 = 0 by graphing.
See it
Graph y = x² − 4x − 5. It crosses at (−1, 0) and (5, 0).
Read the roots
The x-coordinates are −1 and 5, so x = −1 or x = 5.
Check −1
(−1)² − 4(−1) − 5 = 1 + 4 − 5 = 0.
Check 5
5² − 4(5) − 5 = 25 − 20 − 5 = 0.
Answer: x = −1 or x = 5
Explore it — zeros or intersections?
Graphing method — zeros
x² − 6x = −8
y = x² − 6x + 8
The same x-values solve both: x = 2 and x = 4.
Select all roots (x-values), or select none for no real roots.
Explain it — why rearrange?
An x-intercept has output zero. To use x-intercepts, make one side of the equation zero first.
Or keep both sides and graph them separately. The two sides are equal wherever their graphs meet.
Worked example
Solve x² − 6x = −8. Compare two graphing methods.
Rearrange
Add 8 to both sides: x² − 6x + 8 = 0.
Zeros method
Graph y = x² − 6x + 8. Its x-intercepts are (2, 0) and (4, 0).
Intersections method
Graph y₁ = x² − 6x and y₂ = −8. They meet at (2, −8) and (4, −8).
Check in the original
2² − 6(2) = 4 − 12 = −8; 4² − 6(4) = 16 − 24 = −8.
Answer: Both methods give x = 2 or x = 4
Graph Detective — predict, reveal, select and check
x² − 5x + 6 = 0
Predict the number of distinct real roots.
How many real roots?
Two crossings: two distinct real roots. One touch: one distinct repeated root. No contact: no real roots.
Try the repeated-root and no-root equations in Graph Detective. Count points, not how many times a number is written.
See it — roots between grid lines
Graphing method — zeros
x² − 2 = 0
Select all roots (x-values), or select none for no real roots.
x² − 2 = 0 → x = ±√2Exact rootsx ≈ −1.41 or x ≈ 1.41Graph-based estimates to two decimal places1.41² − 2 = −0.0119 ≈ 0Rounding means the check is close, not exact
Real-world idea
When does the ball land?
A ball rises from a platform, then falls to the ground. We need the time when its height reaches zero.
The model h(t) = −5t² + 10t + 15 links input time t in seconds to output height h in metres. Ground-level times are roots of h(t) = 0.
See it — time and height
Graphing method — zeros
h(t) = −5t² + 10t + 15
Select all roots (x-values), or select none for no real roots.
Horizontal input: t, time in seconds. Vertical output: h, height in metres. Both algebraic roots are shown, but the flight uses only 0 ≤ t ≤ 3; landing is at 3 seconds.
Worked example
A ball's height is h(t) = −5t² + 10t + 15 metres, t seconds after a throw. When does it reach the ground?
Model
Input t is time in seconds; output h is height in metres. At ground level h = 0, so solve −5t² + 10t + 15 = 0.
Read the graph
The downward-opening parabola has zeros t = −1 and t = 3. These are roots of the equation.
Check both
h(−1) = −5 − 10 + 15 = 0; h(3) = −45 + 30 + 15 = 0.
Interpret
The model starts at t = 0. Negative time is outside this situation. The ball lands at t = 3 seconds; use 0 ≤ t ≤ 3 during the flight.
Answer: 3 seconds after the throw
Two methods, same solutions
x² − 5x + 6 = 0
Graphing method — zeros
Factoring method
(x − 2)(x − 3) = 0
A factor is zero at x = 2
A factor is zero at x = 3
Both methods give the same solutions. Graphing shows where; factoring explains why.
Step 3 of 4
Practice
Graph reference — compare methods as you practise
x² − 5x + 6 = 0
Graphing method — zeros
Factoring method
(x − 2)(x − 3) = 0
A factor is zero at x = 2
A factor is zero at x = 3
Both methods give the same solutions. Graphing shows where; factoring explains why.
Build it
Challenge
Apply it
Explain it
Step 4 of 4
Mastery Check
Lesson 3.3 mastery check
6 questions · no hints during the check. Answer them all, then check your score — you can retry as many times as you like.