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C1.5 · Connect factored form with intercepts and sketches

Learn to connect factored form with intercepts and sketches through clear examples and targeted practice.

Ontario Grade 12 Mathematics

Polynomial and Rational Functions

Use factors to locate and interpret a polynomial’s intercepts

Factored form makes some graph features easy to read. Each factor can show where a polynomial meets the x-axis. The exponent on that factor helps describe how the graph behaves at the intercept. Substituting zero gives the y-intercept. Together, these clues support a useful sketch without determining every point on the curve.

What you will learn

1. Prerequisite bridge: intercepts and factors

An intercept is a point where a graph meets an axis. An x-intercept lies on the x-axis, so its y-coordinate is zero. A y-intercept lies on the y-axis, so its x-coordinate is zero.
A factor is an expression multiplied by another expression. For example, in f(x)=(x−2)(x+1)f(x)=(x-2)(x+1), the factors are x−2x-2 and x+1x+1. A product equals zero if at least one factor equals zero. This is called the zero-product property.
To find x-intercepts from factored form, set each factor equal to zero. A factor of the form x−ax-a gives the zero x=ax=a, and therefore the intercept (a,0)(a,0). To find the y-intercept, substitute x=0x=0 and write the result as (0,f(0))(0,f(0)).
f(a)=0f(a)=0

2. What a factor’s exponent tells you

A zero is an input that makes the function value zero. Each real zero shown by a factor gives an x-intercept. For example, the factor x−3x-3 gives the zero x=3x=3 and the intercept (3,0)(3,0).
The exponent on a factor is called its multiplicity. At an intercept with odd multiplicity, the graph crosses the x-axis. At an intercept with even multiplicity, it touches the axis and turns back. The exponent tells you this local behaviour; it does not give the exact shape of the whole graph.
For example, (x−a)2(x-a)^2 has an even exponent, so the graph touches and turns at (a,0)(a,0). The factor (x−b)3(x-b)^3 has an odd exponent, so the graph crosses at (b,0)(b,0).
(x−a)m=0(x-a)^m=0

3. Connect the clues to a sketch

A sketch is a drawing that shows the important known features of a graph. Start by marking the x-intercepts from the factors. At each one, note whether the curve crosses or touches the axis. Then substitute x=0x=0 to find and mark the y-intercept.
If you want to check whether the graph is above or below the x-axis between intercepts, choose an input in that interval and evaluate the function. A positive value places the graph above the axis at that input; a negative value places it below.
Draw a smooth curve that fits the intercepts and the crossing or touching behaviour. These clues do not specify every point or the exact location of every bend, so do not treat a sketch as an exact plot.

Factor exponent and behaviour at the x-intercept

FactorZeroMultiplicityGraph behaviour
(x−a)1(x-a)^1x=ax=aOddCrosses the axis
(x−a)2(x-a)^2x=ax=aEvenTouches and turns
(x−a)3(x-a)^3x=ax=aOddCrosses the axis
(x−a)4(x-a)^4x=ax=aEvenTouches and turns

Worked example

Sketching from factored form

Sketch f(x)=−(x+2)2(x−1)f(x)=-(x+2)^2(x-1) by identifying its intercepts and its behaviour at each x-intercept.
  1. Find the x-intercepts
    Set each distinct factor equal to zero. The factor x+2x+2 gives x=−2x=-2, and the factor x−1x-1 gives x=1x=1. Write each as a point on the x-axis.
    (−2,0),(1,0)(-2,0), (1,0)
  2. Read the behaviour at each intercept
    The factor x+2x+2 is squared, so its zero has even multiplicity. The graph touches the axis and turns at (−2,0)(-2,0). The factor x−1x-1 has exponent one, so the graph crosses the axis at (1,0)(1,0).
    (x+2)2,(x−1)1(x+2)^2, (x-1)^1
  3. Find the y-intercept
    Substitute x=0x=0. The function value is four, so the y-intercept is (0,4)(0,4).
    f(0)=−(2)2(−1)=4f(0)=-(2)^2(-1)=4
  4. Check the graph’s position
    Choose x=−1x=-1, which lies between the two x-intercepts. The function value is positive, so the graph is above the x-axis at x=−1x=-1. Use this point along with the intercepts to guide the curve.
    f(−1)=−(1)2(−2)=2f(-1)=-(1)^2(-2)=2
Answer: The sketch has x-intercepts (−2,0)(-2,0) and (1,0)(1,0), and a y-intercept (0,4)(0,4). It touches and turns at x=−2x=-2, crosses at x=1x=1, and is above the x-axis at x=−1x=-1.
Check: The intercepts follow directly from the factors. Substituting x=0x=0 gives 44, and substituting x=−1x=-1 gives 22, so both values are consistent with the stated points.

Common mistakes and how to avoid them

Reading the factor x+2x+2 as giving the zero x=2x=2.
Correction: Set the factor equal to zero. Solving x+2=0x+2=0 gives x=−2x=-2, so the intercept is (−2,0)(-2,0).
Assuming that every x-intercept is a crossing.
Correction: Check the exponent on the factor. Odd multiplicity means crossing; even multiplicity means touching and turning.
Using an x-intercept as the y-intercept.
Correction: For the y-intercept, substitute x=0x=0. The point is (0,f(0))(0,f(0)).
Drawing a curve that ignores one of the intercepts.
Correction: Mark all x-intercepts and the y-intercept first. Then draw a curve consistent with those points and each crossing or touch.

Lesson summary

Check your understanding

Question 1

For g(x)=(x−4)2(x+1)g(x)=(x-4)^2(x+1), what happens at the x-intercept where x=4x=4?
  1. The graph crosses the axis.
  2. The graph touches the axis and turns.
  3. There is no x-intercept at x=4x=4.
  4. The point is a y-intercept.
Show answer and explanation
The graph touches the axis and turns.
The factor (x−4)2(x-4)^2 gives the zero x=4x=4 with even multiplicity. The graph touches and turns at (4,0)(4,0).

Question 2

For h(x)=2(x+3)(x−2)3h(x)=2(x+3)(x-2)^3, what is the y-intercept?
  1. (0,−48)(0,-48)
  2. (0,48)(0,48)
  3. (0,−12)(0,-12)
  4. (0,12)(0,12)
Show answer and explanation
(0,−48)(0,-48)
Substitute x=0x=0: h(0)=2(3)(−2)3=−48h(0)=2(3)(-2)^3=-48. Therefore, the y-intercept is (0,−48)(0,-48).

Question 3

For p(x)=(x+2)(x−5)2p(x)=(x+2)(x-5)^2, which statement correctly describes the x-intercepts?
  1. The graph crosses at (−2,0)(-2,0) and crosses at (5,0)(5,0).
  2. The graph touches at (−2,0)(-2,0) and crosses at (5,0)(5,0).
  3. The graph crosses at (−2,0)(-2,0) and touches at (5,0)(5,0).
  4. The graph touches at both intercepts.
Show answer and explanation
The graph crosses at (−2,0)(-2,0) and touches at (5,0)(5,0).
The factor x+2x+2 has exponent one, so the graph crosses at (−2,0)(-2,0). The factor (x−5)2(x-5)^2 has even multiplicity, so the graph touches and turns at (5,0)(5,0).

Key terms

Factor
An expression multiplied by another expression to form a product.
Zero
An input value that makes the function value equal to zero.
Multiplicity
The exponent on the factor that gives a zero.
Sketch
A drawing that shows important known features of a graph without claiming to plot every point exactly.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation C1.5. It is a study resource, not an official curriculum publication.

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