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Q11 · Connect quadratic factors to graph zeros

Learn to connect quadratic factors to graph zeros through clear examples and targeted practice.

Ontario Grade 10 Mathematics

Quadratic Relations

Reading the x-intercepts of a parabola straight from its factored form

You already know how to expand and simplify expressions like (x+3)(x−5)(x + 3)(x - 5) from your Grade 9 course. In this lesson you will run that process in reverse: you will start with the factored form of a quadratic and read off exactly where its parabola crosses the x-axis. Those crossing points are called zeros, and connecting them to the factors is one of the most powerful shortcuts in Grade 10 math. By the end of this lesson you will see that a factored quadratic gives you the graph's x-intercepts almost for free — no graphing technology required.

What you will learn

Grade 9 Bridge: Functions, Outputs, and the x-axis

A quadratic function is a rule that takes an input, called xx, and produces an output, often written yy or f(x)f(x). When you substitute a number for xx, you get a single number back. For example, if f(x)=x2−4f(x) = x^2 - 4, then f(3)=32−4=5f(3) = 3^2 - 4 = 5.
The graph of a quadratic is a U-shaped curve called a parabola. Every point on the parabola has coordinates (x,y)(x, y), where yy is the output of the function for that xx. The x-axis is the horizontal line where y=0y = 0. Any point where the parabola touches or crosses the x-axis has an output of exactly zero.
Those special x-values are the focus of this lesson. Finding them tells you where the parabola meets the ground, so to speak, and the factored form of the quadratic makes finding them straightforward.

What a Zero Actually Is

A zero of a function is any value of xx that makes the output equal to zero. Formally, x=rx = r is a zero when f(r)=0f(r) = 0. The word 'zero' here refers to the output being zero, not the input.
Consider f(x)=(x−2)(x+5)f(x) = (x - 2)(x + 5). Substitute x=2x = 2: you get f(2)=(2−2)(2+5)=0 × 7=0f(2) = (2 - 2)(2 + 5) = 0 \ \times \ 7 = 0. The output is zero because the first factor becomes zero. Now substitute x=−5x = -5: f(−5)=(−5−2)(−5+5)=(−7)(0)=0f(-5) = (-5 - 2)(-5 + 5) = (-7)(0) = 0. The output is zero because the second factor becomes zero.
This is the key idea: when a product of two numbers equals zero, at least one of them must be zero. This rule is called the Zero Product Property, and it is why factored form makes zeros so easy to find.
f(x)=0f(x) = 0

Reading Zeros from Factored Form

The factored form of a quadratic looks like f(x)=a(x−r)(x−s)f(x) = a(x - r)(x - s), where aa, rr, and ss are numbers. The value aa is the leading coefficient and controls whether the parabola opens up or down and how wide it is. The values rr and ss are the zeros, because substituting either one makes a factor equal to zero.
Notice the sign pattern carefully. The factor (x−r)(x - r) equals zero when x=rx = r. If the factor is written (x+3)(x + 3), rewrite it as (x−(−3))(x - (-3)) to see that r=−3r = -3. In other words, the zero is the opposite sign of the number inside the bracket.
Once you have both zeros, you automatically know two points on the graph: (r,0)(r, 0) and (s,0)(s, 0). These two points sit on the x-axis, one on each side of the parabola's turning point (called the vertex).
f(x)=a(x−r)(x−s)f(x) = a(x - r)(x - s)

Sketching the Parabola Using the Zeros

The two zeros give you the left and right x-intercepts of the parabola. The axis of symmetry — the vertical mirror line of the parabola — always passes halfway between the two zeros. You find it by averaging: x=r+s2x = \frac{r + s}{2}.
Once you know the axis of symmetry, substitute that x-value into the function to find the y-coordinate of the vertex. Together, the two zeros and the vertex give you enough information to draw a clear sketch without plotting many points.
The value of aa tells you the direction: if a>0a > 0 the parabola opens upward (a 'happy' U), and if a<0a < 0 it opens downward (a 'sad' U). Changing the size of aa makes the parabola narrower or wider but does not move the zeros. Changing rr or ss, on the other hand, shifts the zeros left or right along the x-axis.
x=r+s2x = \frac{r + s}{2}

Why the Connection Matters

Connecting factors to zeros lets you move fluently between algebra and graphs. If someone gives you a factored quadratic, you can immediately name the x-intercepts and sketch the curve. If someone gives you the zeros of a parabola, you can write a factored equation for it.
For example, if you know a parabola has zeros at x=1x = 1 and x=6x = 6, a possible equation is f(x)=(x−1)(x−6)f(x) = (x - 1)(x - 6). You could multiply by any non-zero constant aa and still have a valid equation with the same zeros, because multiplying all outputs by aa stretches the curve but leaves the x-intercepts unchanged.
This two-way connection — factor to zero, zero to factor — is the foundation for solving quadratic equations by factoring, a core skill you will use throughout the rest of MPM2D and beyond.

How the Factored Form Reveals Key Graph Features

Factored FormZeros (x-intercepts)Axis of SymmetryOpens Up or Down
f(x)=(x−1)(x−7)f(x) = (x - 1)(x - 7)x=1x = 1 and x=7x = 7x=4x = 4Up (a=1>0a = 1 > 0)
f(x)=(x+2)(x−6)f(x) = (x + 2)(x - 6)x=−2x = -2 and x=6x = 6x=2x = 2Up (a=1>0a = 1 > 0)
f(x)=−3(x+4)(x+1)f(x) = -3(x + 4)(x + 1)x=−4x = -4 and x=−1x = -1x=−2.5x = -2.5Down (a=−3<0a = -3 < 0)
f(x)=2(x−3)(x−3)f(x) = 2(x - 3)(x - 3)x=3x = 3 (one zero, touches axis)x=3x = 3Up (a=2>0a = 2 > 0)

Worked example

Finding Zeros and Sketching from Factored Form

A quadratic function is given in factored form as f(x)=2(x+1)(x−5)f(x) = 2(x + 1)(x - 5). Find the zeros of the function, state the axis of symmetry, and describe the direction the parabola opens.
  1. Identify the factors
    The function has two linear factors: (x+1)(x + 1) and (x−5)(x - 5). The leading coefficient is a=2a = 2.
    f(x)=2(x+1)(x−5)f(x) = 2(x + 1)(x - 5)
  2. Set the first factor equal to zero
    Apply the Zero Product Property. The output is zero whenever either factor is zero. Set (x+1)=0(x + 1) = 0 and solve for xx.
    x+1=0 ⇒ x=−1x + 1 = 0 \ \Rightarrow \ x = -1
  3. Set the second factor equal to zero
    Now set (x−5)=0(x - 5) = 0 and solve for xx.
    x−5=0 ⇒ x=5x - 5 = 0 \ \Rightarrow \ x = 5
  4. State the zeros
    The zeros are x=−1x = -1 and x=5x = 5. These are the x-coordinates where the parabola crosses the x-axis, giving the points (−1,0)(-1, 0) and (5,0)(5, 0).
  5. Find the axis of symmetry
    Average the two zeros to find the x-value of the axis of symmetry.
    x=−1+52=42=2x = \frac{-1 + 5}{2} = \frac{4}{2} = 2
  6. Determine the direction
    The leading coefficient is a=2a = 2, which is greater than zero, so the parabola opens upward.
    a=2>0a = 2 > 0
Answer: Zeros: x=−1x = -1 and x=5x = 5. Axis of symmetry: x=2x = 2. The parabola opens upward.
Check: Verify by substituting each zero back into f(x)f(x): f(−1)=2(−1+1)(−1−5)=2(0)(−6)=0f(-1) = 2(-1+1)(-1-5) = 2(0)(-6) = 0 ✓ and f(5)=2(5+1)(5−5)=2(6)(0)=0f(5) = 2(5+1)(5-5) = 2(6)(0) = 0 ✓.

Worked example

Writing a Factored Equation from Given Zeros

A parabola crosses the x-axis at x=−4x = -4 and x=3x = 3, and it passes through the point (1,−15)(1, -15). Write an equation for this parabola in factored form.
  1. Use the zeros to write the skeleton factored form
    Because the zeros are x=−4x = -4 and x=3x = 3, the factors are (x−(−4))=(x+4)(x - (-4)) = (x + 4) and (x−3)(x - 3). Include the unknown leading coefficient aa.
    f(x)=a(x+4)(x−3)f(x) = a(x + 4)(x - 3)
  2. Substitute the known point to find aa
    The parabola passes through (1,−15)(1, -15), meaning when x=1x = 1 the output is −15-15. Substitute both values.
    −15=a(1+4)(1−3)-15 = a(1 + 4)(1 - 3)
  3. Simplify the right side
    Calculate each factor at x=1x = 1: (1+4)=5(1 + 4) = 5 and (1−3)=−2(1 - 3) = -2. Their product is 5×(−2)=−105 \times (-2) = -10.
    −15=a(−10)-15 = a(-10)
  4. Solve for aa
    Divide both sides by −10-10 to isolate aa.
    a=−15−10=32a = \frac{-15}{-10} = \frac{3}{2}
  5. Write the final equation
    Substitute a=32a = \frac{3}{2} back into the factored form.
    f(x)=32(x+4)(x−3)f(x) = \frac{3}{2}(x + 4)(x - 3)
Answer: f(x)=32(x+4)(x−3)f(x) = \frac{3}{2}(x + 4)(x - 3)
Check: Check the given point: f(1)=32(1+4)(1−3)=32(5)(−2)=32(−10)=−15f(1) = \frac{3}{2}(1+4)(1-3) = \frac{3}{2}(5)(-2) = \frac{3}{2}(-10) = -15 ✓. Check the zeros: f(−4)=32(0)(−7)=0f(-4) = \frac{3}{2}(0)(-7) = 0 ✓ and f(3)=32(7)(0)=0f(3) = \frac{3}{2}(7)(0) = 0 ✓.

Common mistakes and how to avoid them

Copying the number inside the bracket as the zero without flipping the sign. For example, reading (x+4)(x + 4) as the zero x=4x = 4 instead of x=−4x = -4.
Correction: Rewrite every factor as (x−r)(x - r) and the zero is rr. Since (x+4)=(x−(−4))(x + 4) = (x - (-4)), the zero is x=−4x = -4.
Forgetting that the leading coefficient aa does not affect the zeros. Students sometimes try to set a=0a = 0 or use aa when solving for the zeros.
Correction: Only the linear factors determine the zeros. Set each factor equal to zero independently and ignore aa during that step.
Claiming the zero is the full point (r,0)(r, 0) but then plotting it at (0,r)(0, r) on the graph, mixing up x and y.
Correction: Zeros are x-values, so they appear on the horizontal axis. The y-coordinate at an x-intercept is always zero, giving the point (r,0)(r, 0), not (0,r)(0, r).
When asked to write a factored equation from zeros, writing f(x)=(x−r)(x−s)f(x) = (x - r)(x - s) without checking whether a coefficient aa is needed, and missing a given point.
Correction: Always include the coefficient aa in the skeleton form f(x)=a(x−r)(x−s)f(x) = a(x - r)(x - s), then substitute the extra point to solve for aa before finalising the equation.
Averaging the zeros incorrectly when finding the axis of symmetry, for example subtracting instead of adding.
Correction: The axis of symmetry is x=r+s2x = \frac{r + s}{2}, which requires adding the two zeros and dividing by 2, not subtracting them.

Lesson summary

Check your understanding

Question 1

What are the zeros of f(x)=(x−7)(x+2)f(x) = (x - 7)(x + 2)?
  1. x=7x = 7 and x=2x = 2
  2. x=−7x = -7 and x=2x = 2
  3. x=7x = 7 and x=−2x = -2
  4. x=−7x = -7 and x=−2x = -2
Show answer and explanation
x=7x = 7 and x=−2x = -2
Set each factor to zero: x−7=0x - 7 = 0 gives x=7x = 7, and x+2=0x + 2 = 0 gives x=−2x = -2. The factor (x+2)(x + 2) means (x−(−2))(x - (-2)), so the zero is −2-2, not +2+2.

Question 2

A parabola has zeros at x=−3x = -3 and x=9x = 9. What is its axis of symmetry?
  1. x=3x = 3
  2. x=6x = 6
  3. x=−6x = -6
  4. x=12x = 12
Show answer and explanation
x=3x = 3
Average the two zeros: x=−3+92=62=3x = \frac{-3 + 9}{2} = \frac{6}{2} = 3. The axis of symmetry is x=3x = 3.

Question 3

Which factored equation has zeros at x=1x = 1 and x=−6x = -6?
  1. f(x)=(x+1)(x−6)f(x) = (x + 1)(x - 6)
  2. f(x)=(x−1)(x+6)f(x) = (x - 1)(x + 6)
  3. f(x)=(x−1)(x−6)f(x) = (x - 1)(x - 6)
  4. f(x)=(x+1)(x+6)f(x) = (x + 1)(x + 6)
Show answer and explanation
f(x)=(x−1)(x+6)f(x) = (x - 1)(x + 6)
For zero x=1x = 1, the factor must be (x−1)(x - 1). For zero x=−6x = -6, the factor must be (x−(−6))=(x+6)(x - (-6)) = (x + 6). So the equation is f(x)=(x−1)(x+6)f(x) = (x - 1)(x + 6).

Question 4

A parabola has zeros at x=2x = 2 and x=8x = 8 and passes through (5,−9)(5, -9). What is the value of aa in f(x)=a(x−2)(x−8)f(x) = a(x - 2)(x - 8)?
  1. a=1a = 1
  2. a=−1a = -1
  3. a=3a = 3
  4. a=−3a = -3
Show answer and explanation
a=1a = 1
Substitute (5,−9)(5, -9): −9=a(5−2)(5−8)=a(3)(−3)=−9a-9 = a(5-2)(5-8) = a(3)(-3) = -9a. Dividing both sides by −9-9 gives a=1a = 1.

Key terms

Zero (of a function)
An x-value that makes the function's output equal to zero; also called an x-intercept or root.
Factored form
A way of writing a quadratic as a product of factors, such as f(x)=a(x−r)(x−s)f(x) = a(x - r)(x - s).
Zero Product Property
If a product of two numbers equals zero, then at least one of the numbers must be zero.
x-intercept
A point where a graph crosses the x-axis; its coordinates are always of the form (x,0)(x, 0).
Parabola
The U-shaped curve that is the graph of any quadratic function.
Axis of symmetry
The vertical line that divides the parabola into two mirror-image halves; its equation is x=r+s2x = \frac{r + s}{2}.
Leading coefficient
The value aa multiplied in front of the factors in f(x)=a(x−r)(x−s)f(x) = a(x - r)(x - s); it controls direction and width of the parabola.
Vertex
The highest or lowest point of the parabola, located on the axis of symmetry.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 10 Mathematics (MPM2D), study topic Q11. It is a study resource, not an official curriculum publication.

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