A fully factored polynomial contains everything you need to draw a precise sketch, no calculus required. From the leading term you get end behaviour; from the roots and their multiplicities you get intercepts and shape; from $P(0)$ you get the $y$-intercept. This lesson ties all the polynomial tools together into a five-step sketching method.
Describe the end behaviour of $y = x^3$ as $x \to +\infty$ and $x \to -\infty$. Without looking ahead what about $y = -x^4$? How does changing the sign of the leading coefficient change the picture? Can you predict what happens from just the degree and leading coefficient?
Given any polynomial in factored form $P(x) = a(x-\alpha_1)^{m_1}(x-\alpha_2)^{m_2}\cdots$, follow these five steps in order to produce an accurate sketch.
For large $|x|$, the leading term $ax^n$ completely controls $y$. Every other term is tiny by comparison. So end behaviour depends on just two things: degree (even or odd) and sign of $a$.
The four cases for end behaviour. Memorise the pattern: even degree = same ends; odd degree = opposite ends.
End behaviour is determined entirely by the leading term ax^n; Even degree: both ends go the same direction (up if a>0; down if a<0)
Pause, copy the end-behaviour rules into your book: end behaviour is determined by $ax^n$; even degree with $a > 0$: both ends up; even with $a < 0$: both ends down; odd with $a > 0$: left down, right up; odd with $a < 0$: left up, right down.
Quick check: For $y = -2x^3 + 5x^2 - 3$, as $x \to +\infty$:
We just saw that end behaviour is determined by the leading term $ax^n$, both ends go up for positive leading coefficient with even degree, and they go in opposite directions for odd degree. That raises a question: given the fully factored form $P(x) = a(x-\alpha_1)^{m_1} \cdots$, how do you sketch the graph using the roots, multiplicities, and end behaviour together? This card answers it → follow the five-step method: roots, cross/touch behaviour, $y$-intercept, end behaviour, then draw a smooth curve.
Given $P(x) = a(x - \alpha_1)^{m_1}(x - \alpha_2)^{m_2}\cdots$, the sketch follows directly from the five steps:
Five steps: (1) roots, (2) cross/touch, (3) y-int, (4) end behaviour, (5) smooth curve; Never add extra bumps, the degree limits turning points to at most n-1
Pause, copy the five-step sketching method into your book: (1) find roots from factored form; (2) odd multiplicity → cross, even → touch; (3) compute $y$-intercept; (4) determine end behaviour from leading term; (5) draw smooth curve without extra bumps (degree limits turning points to $\le n-1$).
Did you get this? True or false: for an odd-degree polynomial with a positive leading coefficient, the graph falls to the left and rises to the right.
Worked examples · 3 in a row, reveal as you go
Sketch $y = -(x - 1)^2(x + 2)$, showing all intercepts and end behaviour.
Describe the end behaviour of $y = 2x^5 - 3x^2 + 1$.
Sketch $y = (x + 1)^3(x - 2)^2$, showing all intercepts and end behaviour.
Fill the gap: For $y = -(x-3)(x+1)^2(x-2)$, the $y$-intercept is $P(0) = $ and the degree is .
Misconceptions to fix · the traps that cost marks
Did you get this? True or false: a degree-4 polynomial can have at most 3 turning points.
Activities · practice with the ideas
Describe the end behaviour of $y = 2x^5 - 3x^2 + 1$. Use the table format: as $x \to +\infty$, $y \to$ ___; as $x \to -\infty$, $y \to$ ___.
Sketch $y = (x + 1)^3(x - 2)^2$. Show all intercepts, the $y$-intercept, and end behaviour. Describe (in words) the shape at each $x$-intercept.
Find the $y$-intercept of $y = -(x - 3)(x + 1)^2(x - 2)$.
For $y = x^4 - 5x^2 + 4$, factor completely then apply the five-step method to describe the sketch (no diagram needed).
Explain how knowing only the roots and their multiplicities, plus the leading term, is enough to constrain the complete shape of a polynomial graph.
Earlier you were asked about the end behaviour of $y = x^3$ and $y = -x^4$.
$y = x^3$: odd degree, positive $a = 1$ → as $x \to +\infty$, $y \to +\infty$; as $x \to -\infty$, $y \to -\infty$ (down-left, up-right).
$y = -x^4$: even degree, negative $a = -1$ → as $x \to \pm\infty$, $y \to -\infty$ (both ends down).
The key insight: degree tells you whether ends go the same way (even) or opposite ways (odd). The sign of $a$ tells you which direction. Together these two facts, combined with multiplicity at each intercept and the $y$-intercept, fully determine the sketch.
Odd one out: Which of these polynomials has end behaviour where both ends go up?
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Describe the end behaviour of $y = 2x^5 - 3x^2 + 1$. (1 mark)
Q2. Sketch $y = (x + 1)^3(x - 2)^2$, showing all intercepts and end behaviour. (4 marks)
Q3. Find the $y$-intercept of $y = -(x - 3)(x + 1)^2(x - 2)$. (1 mark)
Activity: 1. Leading term $2x^5$; odd + positive: as $x\to+\infty$, $y\to+\infty$; as $x\to-\infty$, $y\to-\infty$ · 2. $x=-1$ (mult 3, odd, flat crossing), $x=2$ (mult 2, even, touching); $P(0)=(1)^3(-2)^2=4$; degree 5, $a=+1$, down-left, up-right · 3. $P(0)=-(0-3)(1)(0-2)=-(−3)(1)(−2)=-6$ · 4. $x^4-5x^2+4=(x-1)(x+1)(x-2)(x+2)$; all roots mult 1, all cross; $P(0)=4$; degree 4, $a=+1$, both ends up · 5. Roots give $x$-intercepts; multiplicities give cross/touch; leading term gives end behaviour, together these constrain every part of the graph.
Q1 (1 mark): Leading term $2x^5$: odd degree, positive $a$. As $x\to+\infty$, $y\to+\infty$; as $x\to-\infty$, $y\to-\infty$ [1].
Q2 (4 marks): $x$-intercepts: $x=-1$ (mult 3, crosses flatly) and $x=2$ (mult 2, touches) [1]. $P(0)=(1)^3(-2)^2=4$, so $y$-int $(0,4)$ [1]. Degree 5, $a=+1$: down-left, up-right [1]. Smooth curve connecting all features correctly shown [1].
Q3 (1 mark): $P(0)=-(0-3)(0+1)^2(0-2)=-(-3)(1)(-2)=-(6)=-6$ [1].
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Next lesson: Polynomial Inequalities using the graph and sign analysis to solve inequalities.
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