A. Characteristics of Functions · Quadratic FunctionsHow Many Zeros? Graphs, Factoring and the Discriminant
A2.1About 65 minutes
Learning goals- I can decide how many zeros a quadratic function has by inspecting its vertex and direction of opening.
- I can use factoring and the discriminant to count zeros.
- I can explain which transformations can change the number of zeros.
- I can choose an efficient strategy to count zeros.
Words to know
discriminant
D = b² − 4ac for ax² + bx + c. Its sign tells the number of real zeros.
double root
A repeated root, as in (x − 3)² = 0. The graph touches the x-axis at its vertex.
real zero
A zero that is a real number, shown on the graph as an x-intercept.
Two, one or none?
A snowball thrown across a field rises and falls. Does its path cross the ground line twice, touch it once, or never reach it at all? For a quadratic function the question is: how many zeros does it have? A parabola can cross the x-axis twice, touch it once (at its vertex), or miss it completely.
The curriculum asks you to answer this question with a variety of strategies: inspecting a graph, factoring, and calculating the discriminant. Each one is quicker in some situations.
Strategy 1: Look at the vertex and the direction
For f(x) = a(x − h)² + k:
- If the parabola opens up (a > 0) and the vertex is below the x-axis (k < 0), it must cross the axis twice.
- If the vertex is on the x-axis (k = 0), there is exactly one zero.
- If the parabola opens up and the vertex is above the axis (k > 0), there are no zeros.
For a < 0 the rule flips: a vertex above the axis gives two zeros. A short way to say it: two zeros when a and k have opposite signs, one when k = 0, none when a and k have the same sign.
Strategy 2: Factor
If f(x) factors into two different linear factors, like (x − 1)(x − 3), there are two zeros. If it is a perfect square, like (2x − 3)², both factors give the same zero, so there is one zero (sometimes called a double root).
Strategy 3: The discriminant
In the quadratic formula x = (−b ± √(b² − 4ac))/(2a), the expression D = b² − 4ac is under the square root.
- D > 0: √D is a positive number, so +√D and −√D give two different zeros.
- D = 0: ±0 gives the same value, so there is one zero, x = −b/(2a), at the vertex.
- D < 0: there is no real square root of a negative number, so there are no real zeros.
Bonus: if D is a perfect square (0, 1, 4, 9, 16, …) and a, b, c are integers, the function factors over the integers.
Worked exampleExample 1: Use the discriminant
How many zeros does f(x) = 3x² − 5x + 4 have?
- a = 3, b = −5, c = 4.
- D = (−5)² − 4(3)(4) = 25 − 48 = −23.
- D < 0, so there are no real zeros. (Check: a > 0 and the vertex is above the axis, since the minimum value is c − b²/(4a) = 4 − 25/12 > 0.)
Worked exampleExample 2: Factor
How many zeros does g(x) = 4x² − 12x + 9 have?
- Notice 4x² = (2x)², 9 = 3² and 12x = 2(2x)(3). It is a perfect square: g(x) = (2x − 3)².
- Both factors give x = 3/2.
- Discriminant check: 144 − 4(4)(9) = 144 − 144 = 0 ✓.
Worked exampleExample 3: Inspect the vertex
How many zeros does h(x) = −2(x + 1)² + 8 have?
- The vertex is (−1, 8), above the x-axis.
- a = −2 < 0, so the parabola opens down from a point above the axis. It must cross the axis twice.
- Check: (x + 1)² = 4 gives x = 1 or x = −3 ✓.
Which transformations change the number of zeros?
Start with f(x) = x² − 4x + 3 = (x − 2)² − 1, which has two zeros (1 and 3). Think about what each transformation from the sibling unit Functions and Equivalent Expressions does to the x-axis crossings:
- Vertical stretch or compression, af(x): every zero still gives a × 0 = 0. Same number of zeros.
- Reflection in the x-axis, −f(x): zeros stay zeros. Same number.
- Horizontal translation f(x − d) or horizontal stretch f(kx): the zeros move or spread out, but there are just as many. Same number.
- Vertical translation f(x) + c: this moves the vertex up or down, so it can change the number of zeros.
Worked exampleExample 4: How far can you shift?
For f(x) = x² − 4x + 3, find the values of c for which g(x) = f(x) + c has two zeros, one zero or no zeros.
- Vertex form: f(x) = (x − 2)² − 1, so g(x) = (x − 2)² − 1 + c. The vertex of g is (2, c − 1), and g opens up.
- Two zeros when the vertex is below the axis: c − 1 < 0, so c < 1.
- One zero when the vertex is on the axis: c = 1.
- No zeros when the vertex is above the axis: c > 1.
- Check with the discriminant: g(x) = x² − 4x + (3 + c), so D = 16 − 4(3 + c) = 4 − 4c. D > 0 exactly when c < 1 ✓.
Two zeros for c < 1, one for c = 1, none for c > 1.
Pick the fastest strategy
Vertex form given? Inspect the vertex. Factored form or an easy trinomial? Factor. Messy standard form? Discriminant. You never need to solve the equation just to count the zeros.
Hands-on activitySlide the parabola
You need: graph paper, a pencil, a clear plastic sleeve or a sheet of tracing paper (baking parchment held up to a window also works) and a dry-erase or washable marker.
- On graph paper, draw the x-axis and y-axis and plot y = x² carefully for x = −3 to 3.
- Trace the parabola onto the plastic or tracing paper.
- Slide the tracing straight up and down over the graph paper (keep the axis of symmetry on the y-axis). Count how many times the traced curve meets the x-axis in each position.
- Now slide it left and right only. Does the count ever change?
- Flip the tracing upside down (reflection) and repeat step 3.
What to look for: sliding up and down changes the count: two zeros when the vertex is below the axis, one when it sits on it, none when it is above. Sliding left and right never changes the count. Flipping reverses which positions give two zeros.
Just for funA parabola with a negative discriminant walked up to the x-axis and said, "I think we should see other planes." (Its zeros really do exist, but only as complex numbers, which you will meet in university math.)
Your turn
1Use the discriminant to find the number of zeros of f(x) = x² + 3x − 10.
D = 9 + 40 = 49 > 0: two zeros. (49 is a perfect square, so it factors: (x + 5)(x − 2).)
2Find the number of zeros of f(x) = 2x² − 4x + 5.
D = 16 − 40 = −24 < 0: no zeros.
3Find the number of zeros of f(x) = 9x² + 6x + 1.
D = 36 − 36 = 0, or (3x + 1)²: one zero, x = −1/3.
4Without calculating, how many zeros does f(x) = −3(x − 2)² − 1 have? Explain.
None. It opens down (a < 0) and its vertex (2, −1) is already below the x-axis.
5How many zeros does f(x) = 0.5(x + 4)² − 2 have? Explain.
Two. It opens up and its vertex (−4, −2) is below the axis.
6How many zeros does f(x) = −x² + 6x − 9 have?
f(x) = −(x² − 6x + 9) = −(x − 3)²: one zero, x = 3.
7f(x) = x² − 6x + 5. How many zeros do g(x) = f(x) + 4 and k(x) = f(x) + 5 have?
f(x) = (x − 3)² − 4. g(x) = (x − 3)²: one zero. k(x) = (x − 3)² + 1: no zeros.
More practice: warm-up
1What does D > 0 tell you about a quadratic function?
It has two different real zeros.
2Find the discriminant of x² + 1 and state the number of zeros.
D = 0 − 4 = −4: no zeros.
3A parabola opens up and has vertex (3, −2). How many zeros?
4How many zeros does f(x) = 5(x − 1)² have?
5Find the discriminant of x² − 5x + 6.
More practice: on your own
1How many zeros does f(x) = −4x² + 3x − 1 have?
D = 9 − 4(−4)(−1) = 9 − 16 = −7: no zeros.
2Show that f(x) = 2x² + 5x − 7 has two zeros, and decide whether it factors over the integers.
D = 25 + 56 = 81, which is positive and a perfect square, so two zeros and it factors: (2x + 7)(x − 1).
3Explain why f(x) = ax² + bx + c always has two zeros when a and c have opposite signs.
Then ac < 0, so −4ac > 0. D = b² − 4ac is b² (never negative) plus a positive number, so D > 0.
4f(x) = x² − 8x + 12. For which values of k does f(x) + k have no zeros? Does f(x − 10) have the same number of zeros as f(x)?
f(x) = (x − 4)² − 4, so f(x) + k has no zeros when k > 4. f(x − 10) is a horizontal translation, so it still has two zeros (at 12 and 16).
5A ball's height is h(t) = −4.9t² + 15t + 1 metres. Use the discriminant to decide whether it reaches 12 m.
Solve −4.9t² + 15t + 1 = 12, or −4.9t² + 15t − 11 = 0. D = 225 − 4(−4.9)(−11) = 225 − 215.6 = 9.4 > 0, so yes: it is at 12 m twice (going up and coming down). Its maximum is about 12.5 m.
Challenge
1Which of these transformations of y = f(x) can change the number of zeros: 3f(x), f(x + 2), −f(x), f(x) − 5, f(2x)? Explain.
Only f(x) − 5. Multiplying outputs by 3 or −1 keeps zeros as zeros, and f(x + 2) and f(2x) move or squeeze the zeros horizontally without removing any. A vertical translation moves the vertex up or down, which can create or remove crossings.
2Expand f(x) = a(x − h)² + k and show that its discriminant is D = −4ak. What does this tell you?
f(x) = ax² − 2ahx + (ah² + k). D = (−2ah)² − 4a(ah² + k) = 4a²h² − 4a²h² − 4ak = −4ak. So D > 0 exactly when a and k have opposite signs, which is the vertex rule.
3Can a quadratic function with c = 0 ever have no zeros? Explain in two ways.
No. (1) f(0) = 0, so x = 0 is always a zero. (2) D = b² − 4a(0) = b² ≥ 0, so D is never negative.
Quiz
1. How many zeros does f(x) = 2x² + 3x + 5 have?
2. How many zeros does f(x) = −(x − 1)² + 4 have?
3. Which transformation of y = f(x) can change the number of zeros?
4. If the discriminant of a quadratic function is 0, then:
5. Which function has exactly one zero?
6. f(x) = x² + 2x + c has two zeros when:
Big idea: Expectation A2.1: determine the number of zeros of a quadratic function using a variety of strategies (inspecting graphs, factoring, the discriminant), including how transformations affect the number of zeros.
How to help: Give your teen a quadratic and ask "How many zeros, and what is the quickest way to tell?" Ask them to explain why D < 0 means no zeros.
Common difficulties:
- Calculating −4ac with the wrong sign when a or c is negative. Write each step.
- Forgetting the square of a negative b is positive: (−5)² = 25.
- Thinking reflections or stretches change the number of zeros.
- Using the vertex rule but forgetting that the sign of a flips it.