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Grade 12 · MHF4U · Advanced Functions

MHF4U: End Behaviour and Degree

End behaviour is what a polynomial's graph does far to the left and far to the right. Two things decide it, the degree and the sign of the leading coefficient, and nothing else in the equation matters out there.

How It Works

Far from the origin, the leading term swamps every other term. In f(x)=2x4−5x3+x−7f(x) = 2x^4 - 5x^3 + x - 7 at x=100x = 100, the 2x42x^4 term is 200,000,000200{,}000{,}000 and the other three together are about 5,000,0005{,}000{,}000. So the ends of the graph behave like the ends of y=2x4y = 2x^4.

Even degree: both ends go the same way. A positive leading coefficient sends both ends up, like y=x2y = x^2; a negative one sends both down. Odd degree: the ends go opposite ways. Positive means down on the left and up on the right, like y=x3y = x^3; negative flips that, up on the left and down on the right.

Write it with arrow notation: as x→−∞x \to -\infty, y→∞y \to \infty and as x→∞x \to \infty, y→−∞y \to -\infty. Some teachers want quadrants instead: that graph runs from quadrant 2 to quadrant 4. Practise saying it both ways.

In factored form, do not expand. The degree is the sum of the exponents on the factors, and the leading coefficient is the product of the constant out front and the coefficients of xx in each factor. For −2(x+1)2(x−3)(2x+5)-2(x + 1)^2(x - 3)(2x + 5) that is degree 2+1+1=42 + 1 + 1 = 4 and leading coefficient −2×1×1×2=−4-2 \times 1 \times 1 \times 2 = -4.

Running it backwards: a graph with the same end behaviour on both sides has even degree, and opposite ends mean odd degree. The turning points give a lower bound, since a degree nn polynomial has at most n−1n - 1 of them. So a graph with four turning points needs degree at least 5.

The trap on tests is a polynomial written in a messy form, where the leading term is hiding. Find the term of highest degree first, then read its sign.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

  1. Easy

    Describe the end behaviour of f(x)=2x4−5x3+x−7f(x) = 2x^4 - 5x^3 + x - 7.
    Show solution
    1. The degree is 4, which is even, so both ends go the same way.
    2. The leading coefficient is 22, which is positive, so both ends go up.

    Answer: As x→−∞x \to -\infty, f(x)→∞f(x) \to \infty, and as x→∞x \to \infty, f(x)→∞f(x) \to \infty. The graph runs from quadrant 2 to quadrant 1.

  2. Easy

    Describe the end behaviour of g(x)=−3x5+4x2−1g(x) = -3x^5 + 4x^2 - 1.
    Show solution
    1. The degree is 5, which is odd, so the ends go opposite ways.
    2. The leading coefficient −3-3 is negative, so the graph is a flipped y=x5y = x^5: up on the left, down on the right.

    Answer: As x→−∞x \to -\infty, g(x)→∞g(x) \to \infty, and as x→∞x \to \infty, g(x)→−∞g(x) \to -\infty. Quadrant 2 to quadrant 4.

  3. Medium

    State the degree, the leading coefficient and the end behaviour of h(x)=−2(x+1)2(x−3)(2x+5)h(x) = -2(x + 1)^2(x - 3)(2x + 5) without expanding.
    Show solution
    1. Add the exponents: 2+1+1=42 + 1 + 1 = 4. The degree is 4.
    2. Multiply the constant and the xx coefficients: −2×1×1×2=−4-2 \times 1 \times 1 \times 2 = -4. The leading coefficient is −4-4.
    3. Even degree, negative leading coefficient: both ends go down.

    Answer: Degree 4, leading coefficient −4-4. As x→±∞x \to \pm\infty, h(x)→−∞h(x) \to -\infty.

  4. Medium

    A polynomial's graph rises on the far left, falls on the far right, and has four turning points. What can you say about its degree and its leading coefficient?
    Show solution
    1. Opposite ends mean the degree is odd.
    2. Rising on the left and falling on the right is the shape of y=−x3y = -x^3, so the leading coefficient is negative.
    3. Four turning points need degree at least 4+1=54 + 1 = 5, since a degree nn polynomial has at most n−1n - 1 turning points.

    Answer: The degree is odd and at least 5, and the leading coefficient is negative.

  5. Test level

    Find the end behaviour of p(x)=(3−x)(x2+1)(2x−1)2+10x3p(x) = (3 - x)(x^2 + 1)(2x - 1)^2 + 10x^3.
    Show solution
    1. The product's leading term comes from the leading term of each factor: (−x)(x2)(2x)2=(−x)(x2)(4x2)=−4x5(-x)(x^2)(2x)^2 = (-x)(x^2)(4x^2) = -4x^5.
    2. The 10x310x^3 term has degree 3, lower than 5, so it cannot change the leading term.
    3. Degree 5 is odd and the leading coefficient −4-4 is negative: up on the left, down on the right.

    Answer: As x→−∞x \to -\infty, p(x)→∞p(x) \to \infty, and as x→∞x \to \infty, p(x)→−∞p(x) \to -\infty.

Learn It on Martian Lab

Martian Lab teaches this in the Qualitative Analysis lesson of its MHF4U Polynomial Functions unit. Watch it, then take the Polynomial Functions practice tests to see if it stuck.

Watch the Qualitative Analysis lesson free

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