A. Characteristics of Functions · Quadratic FunctionsMaximum and Minimum Values: Partial Factoring and Symmetry
A2.2About 65 minutes
Learning goals- I can find the axis of symmetry by averaging the zeros.
- I can use partial factoring to find two points with the same height.
- I can use x = −b/(2a) to find the vertex of any quadratic function.
- I can choose an efficient method to find a maximum or minimum value.
Words to know
axis of symmetry
The vertical line x = h through the vertex; the parabola is a mirror image on each side.
partial factoring
Factoring only the x² and x terms: ax² + bx + c = x(ax + b) + c. It gives two points with the same y-value, c.
x = −b/(2a)
The x-coordinate of the vertex of f(x) = ax² + bx + c.
A shortcut worth knowing
Completing the square always works, but sometimes you only want the maximum or minimum value, not the whole vertex form. Engineers, coaches and business owners care about the best value, and there are faster roads to it. Every one of them is built on one fact you learned in Grade 10:
A parabola is symmetric. The vertex lies on the axis of symmetry, halfway between any two points with the same height.
Find two such points, average their x-coordinates, and you have the x-coordinate of the vertex. Then substitute to get the maximum or minimum value.
Method 1: Average the zeros
If the function factors, the two zeros have the same height (0). The axis is halfway between them.
For f(x) = x² − 2x − 15 = (x − 5)(x + 3), the zeros are 5 and −3, so the axis is x = (5 + (−3)) ÷ 2 = 1. The minimum value is f(1) = 1 − 2 − 15 = −16.
Method 2: Partial factoring
Many quadratics do not factor, for example f(x) = 3x² − 6x + 5. Partial factoring still works. Factor the x² and x terms only:
f(x) = 3x(x − 2) + 5.
Now look at the part 3x(x − 2). It equals 0 when x = 0 or x = 2. At both of those inputs, f(x) = 0 + 5 = 5. So the points (0, 5) and (2, 5) have the same height. The axis is halfway between: x = (0 + 2) ÷ 2 = 1, and the minimum value is f(1) = 3 − 6 + 5 = 2.
This is the curriculum's own example of how partial factoring helps you find a minimum.
Method 3: The formula x = −b/(2a)
Partial factoring works for every quadratic, so we can do it once with letters:
f(x) = ax² + bx + c = x(ax + b) + c.
The part x(ax + b) is 0 when x = 0 or x = −b/a. Halfway between 0 and −b/a is
x = −b/(2a).
This is the x-coordinate of the vertex of every quadratic function. Substitute it to get the maximum or minimum value. (You found the same result by completing the square with letters in the last lesson's challenge.)
Worked exampleExample 1: Partial factoring with a negative a
Find the maximum value of f(x) = −2x² + 8x + 3.
- Factor −2x from the first two terms: f(x) = −2x(x − 4) + 3.
- −2x(x − 4) = 0 when x = 0 or x = 4, so f(0) = f(4) = 3.
- Axis: x = (0 + 4) ÷ 2 = 2.
- f(2) = −2(4) + 16 + 3 = −8 + 16 + 3 = 11.
- a = −2 < 0, so this is a maximum.
Maximum value 11 at x = 2.
Worked exampleExample 2: Using x = −b/(2a)
A small business models its weekly profit, in dollars, by P(x) = −4x² + 28x − 9, where x is the price of its product in dollars. Find the price that gives the maximum profit and the maximum profit.
- a = −4 and b = 28. x = −b/(2a) = −28 ÷ (2 × (−4)) = −28 ÷ (−8) = 3.5.
- P(3.5) = −4(3.5)² + 28(3.5) − 9 = −4(12.25) + 98 − 9 = −49 + 98 − 9 = 40.
- a < 0, so this is a maximum.
A price of $3.50 gives the maximum weekly profit of $40.
Worked exampleExample 3: A fraction for the axis
Find the minimum value of f(x) = 2x² + 7x − 1. Give an exact answer.
- x = −b/(2a) = −7/4.
- f(−7/4) = 2(49/16) + 7(−7/4) − 1 = 49/8 − 49/4 − 1.
- Common denominator 8: 49/8 − 98/8 − 8/8 = −57/8.
- Decimal check: f(−1.75) = 2(3.0625) − 12.25 − 1 = 6.125 − 13.25 = −7.125 = −57/8 ✓.
Minimum value −57/8 (−7.125) at x = −7/4.
| Method | Use it when | What you get |
|---|
| Average the zeros | the function factors easily | axis, then max/min by substitution |
| Partial factoring | you want a quick axis without the formula | two equal-height points, axis, max/min |
| x = −b/(2a) | any quadratic in standard form | axis, then max/min by substitution |
| Completing the square | you also want vertex form (for sketching or transformations) | the full vertex form a(x − h)² + k |
Two numbers, two jobs
x = h tells you where the maximum or minimum happens (the price, the time, the width). f(h) is the maximum or minimum value (the profit, the height, the area). Word problems usually ask for both, so read carefully.
Hands-on activityFold to find the axis
You need: graph paper (or lined paper with a grid you draw), a ruler and a pencil.
- Make a table of values for f(x) = x² − 2x − 15 for x = −4 to 6, plot the points and join them with a smooth curve.
- Fold the paper so the two zeros (−3, 0) and (5, 0) land on top of each other. Unfold. The crease is the axis of symmetry. Read its equation.
- Find the lowest point on your curve. Does it lie on the crease?
- Now pick any two points of your curve with the same y-value (for example, x = −2 and x = 4). Fold them onto each other. Do you get the same crease?
What to look for: the crease is x = 1 every time, and the lowest point is (1, −16). Any pair of equal-height points gives the same crease, which is exactly why partial factoring works.
Just for funPartial factoring is like finding two friends standing at the same height on a hill: stand exactly halfway between them and you will be right under the top (or right over the bottom).
Your turn
1Find the zeros of f(x) = (x − 4)(x + 8), then the minimum value.
Zeros 4 and −8. Axis x = (4 − 8)/2 = −2. f(−2) = (−6)(6) = −36.
2Use partial factoring to find two points with the same y-value on f(x) = x² + 6x + 1.
f(x) = x(x + 6) + 1: the points (0, 1) and (−6, 1).
3Use your answer to the last question to find the minimum value of f(x) = x² + 6x + 1.
Axis x = (0 − 6)/2 = −3. f(−3) = 9 − 18 + 1 = −8.
4Use partial factoring to find the minimum value of f(x) = 4x² − 12x − 7.
4x(x − 3) − 7: equal heights at x = 0 and x = 3, axis x = 1.5. f(1.5) = 9 − 18 − 7 = −16.
5Use x = −b/(2a) to find the maximum value of f(x) = −x² + 10x + 2.
x = −10/(−2) = 5. f(5) = −25 + 50 + 2 = 27.
6Find the maximum value of f(x) = −0.5x² + 3x + 4 and where it occurs.
x = −3/(2 × (−0.5)) = 3. f(3) = −4.5 + 9 + 4 = 8.5, at x = 3.
7A ball's height is h(t) = −4.9t² + 24.5t + 1.5 metres after t seconds. Find the maximum height.
t = −24.5/(−9.8) = 2.5 s. h(2.5) = −30.625 + 61.25 + 1.5 = 32.125 m (about 32.1 m).
More practice: warm-up
1The zeros of a parabola are 2 and 10. What is the equation of its axis of symmetry?
2The points (−1, 7) and (5, 7) are on a parabola. What is its axis of symmetry?
3Partially factor f(x) = 5x² + 20x + 11.
4Use x = −b/(2a) to find the axis of symmetry of f(x) = 2x² − 12x + 1.
5Does f(x) = −7x² + x + 3 have a maximum or a minimum? How do you know?
A maximum, because a = −7 < 0.
More practice: on your own
1Find the minimum value of f(x) = 5x² + 20x + 11.
5x(x + 4) + 11: equal heights at x = 0 and x = −4, axis x = −2. f(−2) = 20 − 40 + 11 = −9.
2Find the maximum value of f(x) = −3x² + 2x + 1. Give an exact answer.
x = −2/(−6) = 1/3. f(1/3) = −3(1/9) + 2/3 + 1 = −1/3 + 2/3 + 1 = 4/3.
3f(x) = −2(x − 1)(x − 9). Find the maximum value without expanding.
Axis x = (1 + 9)/2 = 5. f(5) = −2(4)(−4) = 32.
4Find the minimum value of f(x) = 2x² − 5x + 1 exactly.
x = 5/4. f(5/4) = 2(25/16) − 25/4 + 1 = 25/8 − 50/8 + 8/8 = −17/8.
5A footbridge arch over a creek is modelled by h(x) = −0.02x² + 1.2x, where x is the horizontal distance in metres from one end and h(x) is the height in metres. Find the maximum height of the arch.
Partial factor: h(x) = −0.02x(x − 60), so equal heights (0) at x = 0 and x = 60. Axis x = 30. h(30) = −0.02(900) + 36 = 18 m.
Challenge
1For f(x) = ax² + bx + c, show that the maximum or minimum value is f(−b/(2a)) = c − b²/(4a).
f(−b/(2a)) = a(b²/(4a²)) + b(−b/(2a)) + c = b²/(4a) − b²/(2a) + c = b²/(4a) − 2b²/(4a) + c = c − b²/(4a).
2The parabola y = x² + bx + 5 has its vertex on the line x = −3. Find b and the minimum value.
−b/2 = −3, so b = 6. f(−3) = 9 − 18 + 5 = −4.
3The points (−2, 11) and (6, 11) are on the graph of f(x) = x² + bx + c, and the minimum value is −5. Find b and c.
Axis x = (−2 + 6)/2 = 2, so −b/2 = 2 and b = −4. Vertex form f(x) = (x − 2)² − 5 = x² − 4x − 1, so c = −1. Check: f(6) = 36 − 24 − 1 = 11 ✓.
Quiz
1. The zeros of a quadratic function are −6 and 2. What is the axis of symmetry?
2. Partial factoring of f(x) = 3x² − 12x + 7 gives which pair of equal-height points?
3. What is the axis of symmetry of f(x) = −2x² + 10x − 3?
4. What is the maximum value of f(x) = −x² + 4x + 1?
5. A profit model has its vertex at (12, 3500). Which sentence is correct?
6. Why does partial factoring find the axis of symmetry?
Big idea: Expectation A2.2 asks for algebraic methods for the maximum or minimum, giving completing the square and averaging the zeros as examples; its sample problem asks how partially factoring helps find a minimum. This lesson gives three alternatives to completing the square, all based on symmetry.
How to help: Ask, "Where does the best value happen, and what is the best value?" These are two different numbers. Have your teen check answers with a quick table of values on either side of the vertex.
Common difficulties:
- Using x = b/(2a) without the negative sign.
- Stopping after finding x and forgetting to substitute.
- Substitution errors with negative numbers: (−3)² = 9, but −3² = −9. Use brackets on the calculator.
- In partial factoring, factoring out x but forgetting the coefficient: 4x² − 12x = 4x(x − 3), not x(x − 3).